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3 years ago

Who want 100 points?

Mathematics

2 answers:

Natasha2012 [34]3 years ago

8 0

Answer:

ill take some lol, thank you soooooooo much. How are you doing?

Step-by-step explanation:

aliya0001 [1]3 years ago

7 0

ey thanks friend

I hope you have a wonderful day

also nice pfp

You might be interested in

128 is 74% of what number

GarryVolchara [31]

Percent means parts out of 100
74%=74/100=0.74

'of' means multiply

so
128 is 74% of what means
128=0.74 times what
divide both sides by 0.74
172.97297297297297297297297297297=what
the number is about 173

5 0

3 years ago

Leon bought 50 backpacks at a sporting store. 20% of the backpacks were pink and 1/2 of the backpacks were blue. How many of the

My name is Ann [436]

20% of 50 is 10

Half of 50 is 25

25 + 10 = 35

50 -35 = 15

15 of the backpacks were neither pink or blue

Hope this helps :)

5 0

3 years ago

The top question plezzzzzzz zzzzzzzzz zzzzzzzzz zzzzzzzzz

zaharov [31]

Use a ruler for this problem. Take a picture of the measurement & then I could help you.

8 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: