Please help this is confusing:(

8th grade math help!!

horsena [70]

Step-by-step explanation:

A = 1/2h(b1 + b2)

2A = h(b1 + b2)

h(b1 + b2) = 2A

h = 2A/(b1 + b2)

5 0

2 years ago

Find the solution of the system of equations.<br> 10x+10y=40 <br> 5x+3y=8

ddd [48]

Answer:

x = -2 & y = 6

Step-by-step explanation:

$10x+10y=40$

$10x+10y+ -10y=40+-10y$

$10x=-10y+40$

$\frac{10x}{10} =\frac{-10y+40}{10}$

$x=-y+4$

$5x+3y=8$

$5(-y+4)+3y=8$

$-2y+20=8$

$-2y+20+-20=8+-20$

$-2y=-12$

$\div2$

$y=6$

$x=-y+4$

$x=-6+4$

$x=-2$

Hope this helps.

7 0

2 years ago

Assume that a sample is used to estimate a population proportion p. find the margin of error e that corresponds to the given sta

Leno4ka [110]

Let p be the proportion. Let c be the given confidence level , n be the sample size.

Given: p=0.3, n=1180, c=0.99

The formula to find the Margin of error is

ME = $z _{\alpha/2} \sqrt{\frac{p*(1-p)}{n}}$

Where z (α/2) is critical value of z.

P(Z < z) = α/2

where α/2 = (1- 0.99) /2 = 0.005

P(Z < z) = 0.005

So in z score table look for probability exactly or close to 0.005 . There is no exact 0.005 probability value in z score table. However there two close values 0.0051 and 0.0049 . It means our required 0.005 value lies between these two probability values.

The z score corresponding to 0.0051 is -2.57 and 0.0049 is -2.58. So the required z score will be average of -2.57 and -2.58

(-2.57) + (-2.58) = -5.15

-5.15/2 = -2.575

For computing margin of error consider positive z score value which is 2.575

The margin of error will be

ME = $z _{\alpha/2} \sqrt{\frac{p*(1-p)}{n}}$

= $2.575 \sqrt{\frac{0.30*(1-0.3)}{1180}}$

= 2.575 * 0.0133

ME = 0.0342

The margin of error is 0.0342

7 0

2 years ago

NO LINKS OR ANSWERING WHAT YOU DON'T KNOW?

Rom4ik [11]

Answer:

4,a

5.d

6.c

plz mark me as brainliest

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

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 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

4 0

3 years ago