please help to solve this in written format

Answer:

1.125 hours

Step-by-step explanation:

Given :

Ben's speed = 3 mi/hr

Time before Amanda starts = 1.5 hours

Amanda's speed = 7 mi/hr

Time before Amanda catches up with Ben

Recall :

Distance = speed * time

Distance already covered by Ben before Amanda starts :

(3 * 1.5) = 4.5

Hence, we can setup the equation :

Ben's distance = Amanda's distance

Let time taken = x

4.5 + 3x = 7x

4.5 = 7x - 3x

4.5 = 4x

x = 4.5 / 4

x = 1.125 hours

1.125 * 60 = 67. 5 minutes

Answer:

where is the statement

Step-by-step explanation:

its incomplete po

Answer:

the answer should be e^3

Step-by-step explanation:

i hope it helps you

Answer:

10^3000

Step-by-step explanation:

1000¹⁰⁰⁰

= (10³)¹⁰⁰⁰

= 10^(3×1000)

= 10³⁰⁰⁰ or 10^3000

Answered by GAUTHMATH

Answer:

second option

Step-by-step explanation:

I'm not sure how to explain but if you really need an explanation please message me

The function that represents the absolute function will be y = -|x + 3| + 3. Then the function is represented by graph A.

The absolute function is also known as the mode function. The value of the absolute function is always positive.

If the vertex of the absolute function is at (h, k). Then the absolute function is given as

f(x) = | x - h| + k

The function is given below.

y =    -x,    if x > -3

y = x + 6,  if x ≤ -3

The value of the functions at x = -3 is calculated as,

y = - (-3)

y = 3

y = -3 + 6

y = 3

The capability that addresses the outright capability will be y = - |x + 3| + 3. Then the capability is addressed by diagram A.

The graph is given below.

More about the absolute function link is given below.

https://brainly.com/question/10664936

#SPJ2

Answer:

2/3

Step-by-step explanation:

[tex]2 \frac{1}{4} : \frac{1}{2}[/tex] = [tex]\frac{9}{4} : \frac{1}{2}[/tex]

[tex]\frac{\frac{9}{4} }{\frac{1}{2} }[/tex] = [tex]\frac{3}{x}[/tex]

3 * 1/2 = 9/4x

3/2 = 9/4 x

x = 3/2 ÷ 9/4 = 3/2 * 4/9 = 12/18 = 6/9 = 2/3

Answer:

Answer:

15x(-9)-7

Step-by-step explanation:

it has been translated

Answer:

D. 750(0.75)ˣ

Step-by-step explanation:

Let the new brightness be L'. Since our initial brightness L₀ reduces by 25 %, we have that L' = L₀ - 25% of L₀

L' = L₀ - 0.25L₀

L' = 0.75L₀

Adding the second screen, the new intensity is L" = L' - 25 % of L'  

L" = L' - 0.25 L'

L" = 0.75L'.

Since L' = 0.75L₀,

L" = 0.75L' = 0.75(0.75L₀) = 0.75²L₀

Adding the third screen, the new intensity is L"' = L'' - 25 % of L''  

L'" = L" - 0.25 L"

L"' = 0.75L".

Since L" = 0.75L' = 0.75²L₀

L"' = 0.75L" = 0.75(0.75²L₀) = 0.75³L₀

So, we see a pattern here.

The intensity after x screens is L = (0.75)ˣL₀

Since L₀ = 750 lumens,

L = 750(0.75)ˣ

Answer:

x=10

m=3

Step-by-step explanation:

The angles are the same since the sides are the same length (isosceles triangle)

55 = 5x+5

Subtract 5

55-5 =5x+5-5

50 = 5x

Divide by 5

50/5 = 5x/5

10=x

The altitude is a perpendicular bisector so

5m-3 = 2m+6

Subtract 2m from each side

5m-3-2m = 2m+6-2m

3m-3 = 6

Add 3 to each side

3m-3 +3 =6+3

3m =9

Divide by 3

3m/3 = 9/3

m =3

Answer:

The price of 1 exercise book is $122.45.

Step-by-step explanation:

This question is solved using a system of equations.

I am going to say that:

x is the price of one exercise book.

y is the price of one textbook.

Total cost of 10 exercise books and 2 textbooks is $1400.00.

This means that:

[tex]10x + 2y = 1400[/tex]

Since we want x:

[tex]2y = 1400 - 10x[/tex]

[tex]y = 700 - 5x[/tex]

One also finds the total cost of 3 textbooks and 30 exercise books is $3000.

This means that:

[tex]3x + 30y = 3000[/tex]

Since [tex]y = 700 - 5x[/tex]

[tex]3x + 30(700 - 5x) = 3000[/tex]

[tex]3x + 21000 - 150x = 3000[/tex]

[tex]147x = 18000[/tex]

[tex]x = \frac{18000}{147}[/tex]

[tex]x = 122.45[/tex]

The price of 1 exercise book is $122.45.

If z ⁷ = 128i, then there are 7 complex numbers z that satisfy this equation.

[tex]z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}[/tex]

[tex]\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}[/tex]

(where n = 0, 1, 2, …, 6)

[tex]\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}[/tex]

[tex]\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)[/tex]

Answer:

x = 2.7

Step-by-step explanation:

The given equation is :

[tex](3x)^2-10=56[/tex]

We need to solve it for x.

It can be rewrite as follows:

[tex]9x^2-10=56[/tex]

Adding 10 to both sides,

[tex]9x^2-10+10=56+10\\\\9x^2=66\\\\x=\sqrt{\dfrac{66}{9}}\\\\x=2.70[/tex]

So, the value of x is equal to 2.7.

Answer:

a) 0.15 / 0.09

b) 0.15 / 1

c) 0.15 / 0.23

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]

Sample of 10:

This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]

[tex]Z = 1.28[/tex]

[tex]Z = 1.28[/tex] has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

[tex]\mu = 266, \sigma = 16[/tex]

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{260 -  266}{16}[/tex]

[tex]Z = -0.375[/tex]

[tex]Z = -0.375[/tex] has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 20[/tex], so:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]

[tex]Z = -1.68[/tex]

[tex]Z = -1.68[/tex] has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 50[/tex], so:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]

[tex]Z = -2.65[/tex]

[tex]Z = -2.65[/tex] has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = 2.42[/tex]

[tex]Z = 2.42[/tex] has a p-value of 0.9922.

X = 256

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = -2.42[/tex]

[tex]Z = -2.42[/tex] has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Answer:

4,a

5.d

6.c

plz mark me as brainliest

Step-by-step explanation:

Answer:

1. A

2. C

3. A

Step-by-step explanation:

all the explanations are In the image above

Answer:  Choice D.  (-7, 14)

Work Shown:

[tex](x,y) \to (x-3, y+7)\\\\(-4,7) \to (-4-3, 7+7)\\\\(-4,7) \to (-7, 14)\\\\[/tex]

The point has moved 3 units to the left and 7 units up. See the diagram below.

Answer: hello the two normal probability curves are missing

answer:

a) equal to

b) greater than

c) equal to

Step-by-step explanation:

a) The mean of the shorter normal curve is equal to The mean of the taller normal curve is

b) The standard deviation of the shorter normal curve is greater than  the standard deviation of the taller normal curve

c) The area under the shorter normal curve is equal to  the area under the taller normal curve

Answer:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Required

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex]

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] becomes

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Multiply by 1

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}\cdot 1[/tex]

Express 1 as

[tex]\frac{y^2}{y^2} = 1[/tex]

So, the expression becomes:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} \cdot \frac{y^2}{y^2}[/tex]

Rewrite as:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} \cdot \frac{\sin^2y}{y^2}[/tex]

In limits:

[tex]\lim_{(x,y) \to (0,0)} \frac{\sin^2y}{y^2} \to 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} *1[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2}[/tex]

Convert to polar coordinates; such that:

[tex]x = r\cos\theta;\ \ y = r\sin\theta;[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{(r\cos\theta)^2 (r\sin\theta;)^2}{(r\cos\theta)^2+2(r\sin\theta;)^2}[/tex]

Expand

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2\cos^2\theta+2r^2\sin^2\theta}[/tex]

Factor out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2(\cos^2\theta+2\sin^2\theta)}[/tex]

Cancel out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

Express [tex]2\sin^2 \theta[/tex] as [tex]\sin^2\theta+\sin^2\theta[/tex]

So:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+\sin^2\theta+\sin^2\theta}[/tex]

In trigonometry:

[tex]\cos^2\theta + \sin^2\theta = 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

Evaluate the limits by substituting 0 for r

[tex]\frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0}{1+\sin^2\theta}[/tex]

Since the denominator is non-zero; Then, the expression becomes 0 i.e.

[tex]\frac{0}{1+\sin^2\theta} = 0[/tex]

So,

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Answer:

hope this will help you a lot

Answer:

3x² + 13x + 4

Step-by-step explanation:

I did the steps in my book

Answer:

1. D. 1

2. B. y=a³/x

3. A. y=1/x

Step-by-step explanation:

too long to give te explanations but they're there in the attachments

Answer:

The equation is [tex]A(d) = 80 - 4d[/tex]

Step-by-step explanation:

Linear function:

A linear function for the amount of money in an account after t days is given by:

[tex]A(d) = A(0) - md[/tex]

In which A(0) is the initial value and m is the daily cost.

Your EZ Pass account begins with $80. It costs you $4/day.

This means that [tex]A(0) = 80, m = 4[/tex]

So

[tex]A(d) = A(0) - md[/tex]

[tex]A(d) = 80 - 4d[/tex]

Answer:

[tex] \frac{7 + 7 \sqrt{5} }{ - 4} [/tex]

Step-by-step explanation:

We would multiply the fraction by its conjugate

( A conjugate is a expression that has the same integer or number values but have different signs) for example

[tex]5x + 2[/tex]

and

[tex]5x - 2[/tex]

ARE Conjugates.

The conjugate of

[tex]1 - \sqrt{5} [/tex]

is

[tex]1 + \sqrt{5} [/tex]

So this means we will multiply the expression by 1 plus sqr root of 5 on the numerator and denominator.

Our new numerator will be

[tex]7 \times (1 + \sqrt{5} ) = 7 + 7 \sqrt{5} [/tex]

We can apply the difference of squares for the denominator.

[tex](x + y)(x - y) = x {}^{2} - {y}^{2} [/tex]

So our denominator will be

[tex]1 - 5 = - 4[/tex]

So our rationalized fraction will be

[tex] \frac{7 + 7 \sqrt{5} }{ - 4} [/tex]

Answer:

10 times

Step-by-step explanation:

Multiply 40 by 28

1120

Multiply 16 by 7

112

Divide the two numbers

You get 10

Hope this helps!

Answer:

4104.807

Step-by-step explanation:

4269÷1.04

=4104.807

Answer:

4104.8

Step-by-step explanation:

To solve this problem, the easiest approach is to set up a proportion. Use the following general format;

[tex]\frac{currenncy_1}{currency_2}=\frac{balance_1}{balance_2}[/tex]

This will allow one to describe the relationship between the different currency values. Substitute in the given information and solve;

[tex]\frac{currenncy_1}{currency_2}=\frac{balance_1}{balance_2}[/tex]

[tex]\frac{canadian}{australian}=\frac{balance_1}{balance_2}[/tex]

[tex]\frac{1}{1.04}=\frac{x}{4269}[/tex]

Cross products,

[tex]\frac{1}{1.04}=\frac{x}{4269}[/tex]

[tex](1)(4269)=(1.04)(x)\\4269=1.04x[/tex]

Inverse operations,

[tex]4269=1.04x\\4104.8\approx x[/tex]

Answer:

Following are the response to the given choice.

Step-by-step explanation:

Please find the complete question in the attached file.

Subjective opinion = Question of opinion.

Therefore this requires only just opinion and we don't have to do any actual calculations.

Does this seem like a great number of girls, a little number of girls, or a decent number of girls to you but if 1,300 babies were born, 5 of whom were females?

This is a small number of beautiful gals, in my honest opinion. We anticipate boys and girls to be produced about the very same frequency, thus I expect some half of them to be females if there are 1 300 newborns. You should have roughly 650 girls if 50% of the infants are girls, but now we only have five. That appears to me to be considerably low. which is your own opinion.

Answer:

[tex]x = 7[/tex]

Step-by-step explanation:

Given

[tex]AC = 3x +3[/tex]

[tex]AB = -1+2x\\[/tex]

[tex]BC =11[/tex]

Required

Find x

We have:

[tex]AC = AB + BC[/tex]

So, we have:

[tex]3x + 3 = -1 + 2x + 11[/tex]

Collect like terms

[tex]3x - 2x = -1 + 11 - 3[/tex]

[tex]x = 7[/tex]

Answer:

20 participants

Step-by-step explanation:

Given

[tex]Conditions = 2[/tex]

[tex]Scores = 20[/tex]

Type: Repeated design

Required

The number of participants (n)

The repeated measure design implies that the test was conducted repeatedly on the same sample size.

Since the score in each test is 20; then:

[tex]n = 20[/tex] --- the number of participants