All categories

4 years ago

Please help !! thank you !

Mathematics

2 answers:

lord [1]4 years ago

8 0

Answer: y= -55, 135

X= -10, -9

Step-by-step explanation:

Here is a picture of my work. With this type of problem there are a multiple ways of finding your x and y values.

mezya [45]4 years ago

5 0

Answer: (-9,125)

When you graph this the intersection point is that.

You might be interested in

HELPPPP ASAPPP!!! PLSSSSS I NEED HELPPP

Stels [109]

Answer:

Question 1= 10 and question 2=162

Step-by-step explanation:

use pemdas and multiply and add and stuff

8 0

3 years ago

Read 2 more answers

Which number is the additive inverse of -5?

Juli2301 [7.4K]

Answer:

Step-by-step explanation:

8 0

3 years ago

Find the product. −2 (3.1) lol another

sergij07 [2.7K]

Answer:

-6.2

Step-by-step explanation:

multiply -2 times 3.1

6 0

3 years ago

Read 2 more answers

What are all the answers to table 1?

guapka [62]

Answer:

The y intercept is 39 and the slope is 3/1

Step-by-step explanation:

6 0

3 years ago

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