(NEED DONE QUICK) Triangle ABC has a midsegment at DF and segment DF is parallel to segment BC.
What is the value of x?

Answer:

Option A is the correct answer

Step-by-step explanation:

Where there is + sign it changes to - and where there is - sign it change to + sign for the additive inverse

Answer:

Its the first one <3

Step-by-step explanation:

Answer:

(3x)/2+14=15-x

2*(3x/2+14)=2*(15-x)

3x+14*2=15*2-x*2

3x-2x=30-28

x=2

Answer:

the best way is to follow the law, sales tax is not subject to discount by the seller.

Step-by-step explanation:

both ways result in the same outcome:

improper way, discount after tax

$50+$5(tax)=$55, after discount $55-$11(20%)=$44

proper way, discount before tax

$50-$10(20%)=$40+$4(tax)=$44

Answer:

use symbolab

Step-by-step explanation:

Answer:

Step-by-step explanation:

The distance between two points or a line segment can be found by using the formula

[tex]d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\[/tex]

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-4, 1) and (2, 9)

The length of segment AB is

[tex] |AB| = \sqrt{ ({ - 4 - 2})^{2} + ({1 - 9})^{2} } \\ = \sqrt{( { - 6})^{2} + ({ - 8})^{2} } \\ = \sqrt{36 + 64} \\ = \sqrt{100} \: \: \: \: \: \: \: \: \\ = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

We have the final answer as

Hope this helps you

Answer:

The length of segment AB is 10

Step-by-step explanation:

The distance formula is d=[tex]\sqrt{ (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}[/tex]

With the points (-4,1) and (2,9) you know that [tex]-4=x_{1}, 2=x_{2}, 1=y_{1} ,9=y_{2}[/tex]

Now you put the numbers into the distance formula

d=[tex]\sqrt{(2-(-4))^{2}+(9-1)^2[/tex]

(2-(-4) turns into (2+4) because two negatives equal a positive.

After adding and subtracting you get d=[tex]\sqrt{6^2+8^2[/tex]

You then square [tex]6^2[/tex] and [tex]8^2[/tex] to get [tex]\sqrt{36+64[/tex]

After adding 34+64 you get [tex]\sqrt{100[/tex], which is 10

So the length of segment AB is 10

The sampling method used in this scenario is c. convenience sampling.

Convenience sampling involves selecting individuals who are readily available and easily accessible. In this case, the hotel manager asked each guest that checked in on May 15th about their preferences. The manager likely approached the guests based on convenience, as they were already present at the hotel.

Convenience sampling is a quick and convenient way to gather information, but it may introduce bias since the sample may not accurately represent the larger population. It's important to note that convenience sampling may not provide a representative sample and may not yield generalizable results.

The other sampling methods are not applicable in this case. Voluntary response sampling is a type of non-probability sampling in which individuals volunteer to participate in a study. Stratified sampling is a type of probability sampling in which the population is divided into groups (strata) and then a random sample is selected from each group. Systematic sampling is a type of probability sampling in which every nth individual is selected from the population.

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Answer:

(Hope I got them correct I may have some mistakes)

1. $4.80 per pound

2. $2.90 per gallon

3. 31 cents or $0.31, 29 cents or $0.29, 2 cents

4. $10.50, 12 hours

5. 40 inches

6. 13 hours and 20 min

Step-by-step explanation:

1. $15.60/3.25=

$4.80 per pound

2. $12.47/4.3=

$2.90 per gallon

3. $2.48/8=

31 cents or $0.31

$3.48/12=

29 cents or $0.29

31 cents- 29 cents= 2 cents

4. $68.25/6hr 30 min=

$68.25/6.5=

$10.50

$126/10.5= 12 hours

5. Hope you are doing a ratio problem

12/32=15/x

15*32= 480

480/12=

40 inches

6. 360/12=

30

400/30=

13 1/3=

13 hours and 20 min

Answer:

B

Step-by-step explanation:

Answer:

x/2-100=5

Step-by-step explanation:

a. The proportion of days with more than 650 hits is approximately 0.0808.

b. The proportion of days with between 580 and 650 hits is approximately 0.6772.

c. The number of hits such that only 15% of the days will have fewer hits is approximately 406.36.

To solve these questions, we will use the standard normal distribution and the z-score.

a. To find the proportion of days with more than 650 hits, we need to calculate the z-score for 650 and find the area under the standard normal curve to the right of that z-score.

First, calculate the z-score:

z = (x - μ) / σ

z = (650 - 510) / 100

z = 1.4

Using the standard normal table or a calculator, we find the area to the right of the z-score 1.4 is approximately 0.0808.

Therefore, the proportion of days with more than 650 hits is 0.0808 (rounded to four decimal places).

b. To find the proportion of days with between 580 and 650 hits, we need to calculate the z-scores for 580 and 650 and find the area between these two z-scores under the standard normal curve.

First, calculate the z-score for 580:

z1 = (580 - 510) / 100

z1 = 0.7

Next, calculate the z-score for 650:

z2 = (650 - 510) / 100

z2 = 1.4

Using the standard normal table or a calculator, we find the area to the right of z1 (0.7) is approximately 0.7580, and the area to the right of z2 (1.4) is approximately 0.0808. To find the area between these two z-scores, we subtract the smaller area from the larger area:

0.7580 - 0.0808 = 0.6772

Therefore, the proportion of days with between 580 and 650 hits is 0.6772 (rounded to four decimal places).

c. To find the number of hits such that only 15% of the days will have fewer hits, we need to find the z-score that corresponds to the cumulative probability of 0.15 (15%).

Using the standard normal table or a calculator, we find that the z-score corresponding to a cumulative probability of 0.15 is approximately -1.0364.

Now, we can use the z-score formula to find the number of hits:

z = (x - μ) / σ

-1.0364 = (x - 510) / 100

Solving for x:

x - 510 = -1.0364 * 100

x - 510 = -103.64

x = 510 - 103.64

x ≈ 406.36

Therefore, the number of hits such that only 15% of the days will have fewer hits is approximately 406.36.

In summary:

a. The proportion of days with more than 650 hits is approximately 0.0808.

b. The proportion of days with between 580 and 650 hits is approximately 0.6772.

c. The number of hits such that only 15% of the days will have fewer hits is approximately 406.36.

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Answer:

x= -80/77

Step-by-step explanation:

-33x+x+8=-10x

-2.3x+8=-10x

-2.3x=-10-8

-2.3x+10x=-8

7.7x=-8

x=-80/77

Answer:

18

X + 12 = 30

-12. -12. Subtract 12

X = 18. Answer

Answer:

42

Step-by-step explanation:

Answer:

Katrina = 0.8

Susan = 0.833

Gabrielle = 0.85

Savannah = 0.9

Step-by-step explanation:

Change to decimal

Gabrielle = 85%

= 85/100

= 17/20

= 0.85

Savannah = 0.9

= 9/10

= 0.9

Katrina = 4/5

= 0.8

Susan = 5/6

= 0.833

Arrange from least to greatest

Katrina = 0.8

Susan = 0.833

Gabrielle = 0.85

Savannah = 0.9

Answer:

Step-by-step explanation:

Step1

Definition permutation (order is important):

P(n,r)= n!/(n−r)!

Definition combination (order is not important):

C(n,r) =( n/r )= n!/r!(n−r)!

Step2

a) The order of the students does not matter, because a different order results in the same students being selected, and thus we should use the definition of combination.

We are selecting 3 of the 17 students in the class (of which 3 are math majors and 14 are not math majors).

# of possible outcomes=(17,3)=17!/3!(17−3)!=17!/3!14!=680

There areC(14,3) ways to select 3 of the 14 no math majors.

# of favorable outcomes=C(14,3)= 14!/3!(14−3)! = 14!/3!11! =364

The probability is the number of favorable outcomes divided by the number of possible outcome.

P(no math majors)= # of favorable outcomes/# of possible outcomes

                             = 364/680= 91/170

                             ≈0.5353=53.53%

Step 3

(b) By part (a):

P(no math majors)=91/170≈0.5353=53.53%

Use the Complement rule

Complement rule:

P(E’)=1−P(E)

P(at least one math major)=1−P(no math majors)=1−91/170

=79/170≈0.4647=46.47%

Step 4

(c) The order of the students does not matter, because a different order results in the same students being selected, and thus we should use the definition of combination.

We are selecting 3 of the 17 students in the class (of which 3 are math majors and 14 are not math majors).

# of possible outcomes=C(17,3)= 17!/3!(17−3)!=17! /3!14!=680

There areC(14,1) ways to select 1 of the 14 no math majors and there are

C(3,2) ways to select 2 of the 3 major majors.

# of favorable outcomes=C(14,1)⋅C(3,2)=14⋅3=42

P(2 math majors)= # of favorable outcomes/# of possible outcomes

                           = 42/680= 21/340=approx0.0618=6.18%

​

Answer:

3⅓×2 1/6

10/3×13/6

5/3×13/3

65/3

21 2/3

Separate the inequalities by part.

[tex]8<12+c\\12+c<12[/tex]

We'll do 8 < 12+c first then 12+c<12 next.

[tex]8-12<c\\-4<c\\c>-4[/tex]

Then 12+c<12

[tex]12+c<12\\c<12-12\\c<0[/tex]

Then mix c<0 and c>-4 as we'll get [tex]-4<c<0[/tex] #

Answer:

-10

Step-by-step explanation:

f(6)=-2(6)+2

f(6)=-12+2

f(6)=-10

Answer:

 f(6) = - 10

Step-by-step explanation:

-2(6)+2

-12+2

= -10

By proving both directions, we have established the equivalence between a language A being Turing recognizable and the existence of a decider V such that for all x ∈ Σ∗, x ∈ A if and only if ∃y ∈ Σ∗ such that V accepts (x, y).

To prove the statement, we need to show both directions of the "if and only if" statement.

First, let's assume that language A is Turing recognizable. This means that there exists a Turing machine M that recognizes A. We will construct a decider V based on M as follows:

V = "On input (x, y):

Simulate M on input x and y.

If M accepts, accept."

Now, let's analyze the behavior of V. If there exists y such that V accepts (x, y), it means that M accepts both x and y as inputs. Since M recognizes language A, this implies that x ∈ A. Therefore, if V accepts (x, y), then x ∈ A.

On the other hand, if x ∈ A, we know that there exists some string y such that M accepts (x, y). In this case, V will also accept (x, y) because it simulates M and accepts when M accepts. Therefore, if x ∈ A, there exists y such that V accepts (x, y).

We have shown that for all x ∈ Σ∗, x ∈ A if and only if ∃y ∈ Σ∗ such that V accepts (x, y). Hence, the existence of the decider V satisfies the condition for language A to be Turing recognizable.

Next, let's assume that there is a decider V such that for all x ∈ Σ∗, x ∈ A if and only if ∃y ∈ Σ∗ such that V accepts (x, y). We will construct a Turing machine M that recognizes A based on V as follows:

M = "On input x:

For each y in Σ∗, do the following:

Simulate V on input (x, y).

If V accepts, accept."

Now, let's analyze the behavior of M. If x ∈ A, then there exists some y such that V accepts (x, y) according to the assumption. In this case, M will eventually simulate V on input (x, y) and accept. Therefore, if x ∈ A, M will accept.

On the other hand, if x ∉ A, it means that for every y in Σ∗, V rejects (x, y). In this case, M will simulate V on all possible pairs (x, y) and reject for each of them. Therefore, if x ∉ A, M will reject.

We have shown that M recognizes language A based on the existence of the decider V.

By proving both directions, we have established the equivalence between a language A being Turing recognizable and the existence of a decider V such that for all x ∈ Σ∗, x ∈ A if and only if ∃y ∈ Σ∗ such that V accepts (x, y).

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Answer:

x≤4

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

−1/4

x≥−1

Step 2: Multiply both sides by 4/(-1).

(4−1)*(−14x)≥(4−1)*(−1)

Answer:

1. multiply both sides of the equality by -4 and flip the equality sign

that makes the problem: -4×x/-4 >_ -4×(-1)

2.since you multiplied -4 the divisor cancels and you are left with : x<_ -4×(-1)

multiplying 2 negatives makes it a positive so now you get x<_4

Answer:

n = 5/2

Step-by-step explanation:

(2n + 4) + 6 = -9 + 4(2n + 1)

2n + 4 + 6 = -9 + 8n + 4

2n + 10 = 8n - 5

15 = 6n

n = 5/2

Hope this helps!

Answer:

uhmmm try to search on google ‍♂️

Because if the exponent is an even number, x can also be -x. For example, -4^2=4^2 but -4^3 isn't the same as 4^3

Given, the coin is tossed three times.

Sample space: {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}

Number of positive outcomes = 8

Probability of

(1) A: Getting at least two heads

P(A) = P(Getting 3 heads) + P(Getting 2 heads)

        = [tex]\frac{3}{8} +\frac{1}{8}[/tex]                           Since, P(event) = [tex]\frac{No. of favourable outcomes}{Total no. of possible outcomes}[/tex]

P(A)  = [tex]\frac{1}{2}[/tex]            

(2) B: Getting exactly two heads

P(B) = [tex]\frac{3}{8}[/tex]

(3) C: Getting at most one head

P(C) = P(Getting zero head) + P(Getting 1 head)

       = [tex]\frac{1}{8} + \frac{3}{8}[/tex]

       = [tex]\frac{4}{8}[/tex]

P(C) = [tex]\frac{1}{2}[/tex]

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The given statement when calculating the standard error for p-hat, we use the p-hat not the p under h null is false.

When calculating the standard error for p-hat (the sample proportion), we use the value of p under the null hypothesis (p₀) rather than the observed sample proportion (p-hat). This is because the standard error is an estimate of the variability or uncertainty in the sample proportion compared to the population proportion assumed under the null hypothesis.

The standard error formula for p-hat is:

SE(p-hat) = sqrt[(p₀ * (1 - p₀)) / n]

Here, p₀ represents the hypothesized population proportion under the null hypothesis, and n is the sample size. We use p₀ because it represents the proportion assumed to be true in the population when there is no evidence to reject the null hypothesis.

Using p-hat (the observed sample proportion) instead of p₀ would not accurately reflect the assumptions made under the null hypothesis and could lead to incorrect inference and conclusions.

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Answer: 15

Step-by-step explanation:

The probability that either event A or event B will occur is approximately 0.9715.

To find the probability that either event A or event B will occur, we can use the formula P(A or B) = P(A) + P(B) - P(A and B). In this case, the probabilities of events A and B are given as 12/15 and 18/21, respectively. To find the probability of both A and B happening together, we can multiply their individual probabilities. Now we can plug these values into the formula to find the probability of A or B occurring.

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 12/15 + 18/21 - (12/15) * (18/21)

P(A or B) = 0.8 + 0.8571 - (0.8) * (0.8571)

P(A or B) = 0.8 + 0.8571 - 0.6856

P(A or B) = 0.9715

Therefore, the probability that either event A or event B will occur is approximately 0.9715.

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Answer:

Sorry, no thanks

Step-by-step explanation:

The answer for the equation is A and B... I think

Answer:

A B that is what I think the answer is

Answer:

n is smaller, n + 1 is larger

Step-by-step explanation:

First, make the equation.

2(n + n+1) = 510

Then, simplify.

2(2n + 1) = 510

Distribute.

4n + 2 = 510

Solve.

n = 127

The smaller number is n (127) and the larger number is n + 1 (128).

Answer:

the correct answer is the smaller number is:127 and the bigger number is:128

Step-by-step explanation: