Ask question

Ask question

All categories

3 years ago

9

Jamie is X years old. His friend Ana is the same age. Enter an expression in the box to represent the sum of Jamie and ana's age

s.

1 answer:

Lerok [7]3 years ago

4 0

Answer:

2x

Step-by-step explanation:

Since Jamie and Ana have the same age, they are both x years old. So x+x=2x.

You might be interested in

I need help please <br>

Triss [41]

She paid $3.28 per gallon.

24.60 divided by 7.5 equals 3.28.

8 0

2 years ago

The histogram shows the ages of people in a community chorus.

Pachacha [2.7K]

The answer would be 31

5 0

3 years ago

Read 2 more answers

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

2 years ago

Use prime factorization to reduce 36/180

Nataly [62]

The fraction 36/180 can be recuced to 1/5 because 36 divided by 36 is 1 and 180 divided by 36 is 5 hope it helped :)

3 0

2 years ago

Write the sentence as an equation.<br><br> 260 is equal to d plus 236 ?

s344n2d4d5 [400]

Answer:

260 = d + 236 or d + 236 = 260

Step-by-step explanation:

Mental Math

7 0

2 years ago