How many grams of the isotope remains after 90 days?

Using the z-distribution, it is found that since the p-value is greater than 0.05, there is not enough evidence to conclude that there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes.

At the null hypothesis, it is tested if the proportion is still of 44%, that is:

[tex]H_0: p = 0.44[/tex]

At the alternative hypothesis, it is tested if the proportion is now different of 44%, that is:

[tex]H_1: p \neq 0.44[/tex]

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

For this problem, the parameters are:

[tex]p = 0.44, n = 1360, \overline{p} = \frac{626}{1360} = 0.4603[/tex]

Hence the test statistic is:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.4603 - 0.44}{\sqrt{\frac{0.44(0.56)}{1360}}}[/tex]

z = 1.51

Using a z-distribution calculator, for a two-tailed test, as we are testing if the proportion is different of a value, with z = 1.51, the p-value is of 0.1310.

Since the p-value is greater than 0.05, there is not enough evidence to conclude that there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes.

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Using the addition rule of the Sine function and the Cosine function, we obtain [tex]\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex].

Given that

[tex]\sin x + \sin y = a\\\cos x + \cos y = b[/tex]

Then using the above formulas, we get:

[tex]2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=a[/tex]       (1)

[tex]2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=b[/tex]       (2)

Dividing the equation (1) by (2), we get:

[tex]\dfrac{\sin\dfrac{x+y}{2}}{\cos\frac{x-y}{2}}=\dfrac{a}{b}\\\Longrightarrow \tan\dfrac{x+y}{2}=\dfrac{a}{b}[/tex]             (3)

Now, we know that  [tex]\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}[/tex].

Thus, using the above formula, we get from (3):

[tex]\cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\tan^2\dfrac{x+y}{2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\dfrac{a^2}{b^2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex]

Therefore, using the addition rule of the Sine function and the Cosine function, we obtain [tex]\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}[/tex].

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Answer:

[tex]\pi[/tex]([tex]r^{2}[/tex]h)

Step-by-step explanation:

r = radius

h = height

Answer:

exact: SA = 260π in.²

approximate: SA = 816.81 in.²

Step-by-step explanation:

SA = 2πr² + 2πrh

r = 10 in.

h = 3 in.

SA = 2πr(r + h)

SA = 2π(10 in.)(10 in. + 3 in.)

SA = 260π in.²

SA = 816.81 in.²

The y- coordinate that divides the directed line segment from J to K into a ratio of 2:3 is 5. Option C

Let's the point that divides the line segment as point S.

We have that,

Point S divides the line segment into ratio 2:3

The ratio 2:3 means that we are to divide the line segment into;

= 2+3

= 5 equal parts.

We then have that the horizontal distance between the two coordinates is 5

The vertical distance between the two coordinates is 10

Now, let's divide both vertical and horizontal distance into five equal parts,

Horizontal distance = 5/5 = 1

Vertical distance = 10/ 5 = 2

The horizontal distance is 1

The vertical distance is 2

For every one unit move to the left from point J and two units up, we are dividing the line segment into five equal parts as shown in the picture.

The coordinate of point S that divides the line segment into 2 parts and 3 parts is (-5,5)

Thus, the y- coordinate that divides the directed line segment from J to K into a ratio of 2:3 is 5. Option C

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Answer:

8√2 cm²

Refer to the attached page,I've shown the complete calculation over there

Answer:

= 5.375

Step-by-step explanation:

Use the algorithm method.

7  9 9 10

8 . 0 0 0

- 2 . 6 2 5

5 . 3 7 5

= 5.375

Answer:

5 3/8

Step-by-step explanation:

change it so that every number is a improper fraction and same denominator

64/8 - 21/8 = 43/8 = 5 3/8

The times interest earned ratio for 20Y8 and 20Y9 are 11 and 12.6 respectively.

The change in the times interest earned ratio is favorable.

According to the question,

In 20Y8 and 20Y9, Averill Products Inc. disclosed the following on its income statement:

For 20Y9,

Interest expense= $440,000

Income before income tax expense= $5,544,000

Times interest earned ratio= 5544000/440000 = 12.6

For 20Y8,

Interest expense=$400,000

Income before income tax expense=$4,400,000

Times interest earned ratio= 4400000/400000 =11

As the times interest earned ratio for 20Y9 is greater than that for 20Y8,the change in the times interest earned ratio is favorable.

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Answer:225 jeans.

Step-by-step explanation: What we need to do here is convert the number of jeans to a percentage. There were 25 jeans when 50 customers were surveyed and 25 is half of 50 or 50%. This means that if 450 pairs of pants are ordered half of them should be jeans. 450/2 = 225.

The functions, q(x) and r(x) are defined as 2•x + 1, and -5•x - 3, respectively, therefore;

[tex] \: (q \: \circ \: r) (1)= - 15[/tex]

[tex] \: (r \: \circ \: q) (1)= -18[/tex]

The given functions are;

q(x) = 2•x + 1

r(x) = -5•x - 3

The evaluation of the composite functions can be presented as follows;

[tex](q \: \circ \: r) (x)= q(r(x))[/tex]

Therefore;

[tex](q \: \circ \: r) (1)= q(r(1))[/tex]

Which gives;

[tex](q \: \circ \: r) (1)= q( - 8)[/tex]

q(-8) = 2×(-8) + 1 = -15

Therefore;

[tex](q \: \circ \: r) (1)= - 15[/tex]

Similarly, we have;

[tex](r \: \circ \: q) (1)= r(q(1))[/tex]

q(1) = 2×1 + 1 = 3

r(3) = -5×3 - 3 = -18

Which gives;

[tex](r \: \circ \: q) (1)= -18[/tex]

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                     _

Answer: 0.83

Step-by-step explanation:

5/6 = 0.8333...

                        _

0.8333... = 0.83

The count of the denominations of the bills of $1, $5, and $10, are 15, 10, and 5, respectively.

In the question, we are given that there are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5-bills than $1-bills. There are half as many $10-bills as $5-bills.

We are asked to find the count of each denomination, given there was altogether $115 in the bag.

We assume the number of $1-bills in the bag to be x.

Total amount in x bills of $1 =  x * $1 = $x.

Given that there are 5 fewer $5-bills than $1-bills in the bag, number of $5-bills in the bag = x - 5.

Total amount in (x - 5) bills of $5 = (x - 5) * $5 = $5(x - 5).

Given that there are half as many $10-bills as $5-bills in the bag, number of $10-bills in the bag = (x - 5)/2.

Total amount in (x - 5)/2 bills of $10 = (x - 5)/2 * $10 = $5(x - 5).

Thus, the total amount in the bag = $x + $5(x - 5) + $5(x - 5).

But, the total amount in the bag = $115.

Thus, we get an equation:

$x + $5(x - 5) + $5(x - 5) = $115,

or, x + 5x - 25 + 5x - 25 = 115,

or, 11x = 115 + 50,

or, 11x = 165,

or, x = 165/11,

or, x = 15.

Thus, number of $1-bills = x = 15.

The number of $5-bills = x - 5 = 15 - 5 = 10.

The number of $10-bills = (x - 5)/2 = (15 - 5)/2 = 10/2 = 5.

Thus, the count of the denominations of the bills of $1, $5, and $10, are 15, 10, and 5, respectively.

The provided question is incomplete. The complete question is:

There are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5-bills than $1-bills. There are half as many $10-bills as $5-bills. If there is $115 altogether, find the number of each type of bill in the wallet."

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