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2 years ago

If 65% of Americans are overweight, how many people would be overweight in a city of 1.5 million?

Mathematics

2 answers:

Harlamova29_29 [7]2 years ago

6 0

Answer:

hey There

Step-by-step explanation:

This compares with 27.5 percent of those age 55 to 64 and 25.6 percent of those age 65 to 74. With respect to being overweight, 31.8 percent of the individuals 75 and older were such, while 37.9 percent of 55 to 64 year olds and 37.8 percent of individuals age 65 to 74 were overweight (figure 1).

Sholpan [36]2 years ago

4 0

Answer:

66.9%

i hope thats right

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What if the following choices will simplify 7-5(3)^2

rewona [7]

Answer:

-38

Step-by-step explanation:

7-5(3)^2

PEMDAS

Since there are no parentheses to execute

Exponents next

7-5(3)^2

7-5*9

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7 - 45

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2 years ago

Evaluate the function for x = 2 and x = 6 f(x) = -(x - 2)

12345 [234]

Answer:

f(2) = 0 and f(6) = -4

Step-by-step explanation:

First, find f(x) when x = 2

Plug in 2 as x in the function:

f(x) = -(x - 2)

f(2) = -(2 - 2)

f(2) = -(0)

f(2) = 0

Next, find f(x) when x = 6. Plug in 6 as x in the function:

f(x) = -(x - 2)

f(6) = -(6 - 2)

f(6) = -(4)

f(6) = -4

So, f(2) = 0 and f(6) = -4

3 0

3 years ago

Select the correct answer.

Colt1911 [192]

Answer: B

Step-by-step explanation:

An individual's effective tax rate is the average tax rate.

7 0

2 years ago

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: