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3 years ago

What is the mean of -3,-8,12,-15,9

1 answer:

6 0

Mean = sum of the observations
--------------------------------
<span> number of observations
</span>=> -3 + (-8) + 12 +(-15)+ 9
-----------------------------------
5
=> mean = -1 ..

Hope it helps!!!!

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Y=2x-4<br> What are the five ordered pairs

yarga [219]

Answer:

y= 2x+9

y=2x-1

y=2x-13

y=2x-2

y=2x+7

6 0

2 years ago

Divided 80 by my number, and then added 13. The result was 93. What was my number?

igor_vitrenko [27]

The number is 6400.

We have given that,

divided 80 by my number, and then added 13.

<h3>What is the expression?</h3>

An expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Suppose the number is x

$\frac{x}{80}+13=93$

Subtract 13 from both sides

$\frac{x}{80}+13-13=93-13$

$\frac{x}{80}=80$

Multiply both sides by 80

$\\\frac{80x}{80}=80\cdot \:80$

x=6400

Therefore the number is 6400.

To learn more about the number visit:

brainly.com/question/251701

#SPJ1

4 0

2 years ago

Two streets intersect at a 30 degree angle. At the intersection, there are four crosswalks formed that are the same length. what

ELEN [110]

Answer:

rhombus

Step-by-step explanation:

A quadrilateral with equal-length sides is a <em>rhombus</em>.

_____

A square is a special case of a rhombus.

3 0

3 years ago

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

Find the minimum and maximum values of the objective function z=3x+5y

Margarita [4]

3.15 addd all of them! Then divide

6 0

3 years ago

Read 2 more answers