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Black_prince [1.1K]
3 years ago
13

What 2 square numbers have a difference of 55

Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0
a^2-b^2 =55 \\  \\ 28^2 -27^2=55 \\  \\ 28^2 =784 \\  \\27^2 =729 \\  \\ 784-729=55 \\  \\  \\ \sf So,~the~ difference~ of~ the~ two~ squares~ of~ 28~ and~ 27~ equals~ to~ 55.
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Elis [28]

Answer:

D) y=5x+20

Step-by-step explanation:

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3 years ago
A doctor released the results of clinical trials for a vaccine to prevent a particular disease. In these clinical​ trials, 200,0
CaHeK987 [17]

Answer:

1) No, there is  not  enough evidence to support the claim that the vaccine was effective (P-value=0.104) .

The null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0

being the subindex 1 for the experimental group and subindex 2 for the control group.

2) Test statistic z=-1.26

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the vaccine was effective.

Then, the null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0

The significance level is 0.01.

The sample 1 (experimental group), of size n1=100000 has a proportion of p1=0.0002.

p_1=X_1/n_1=21/100000=0.0002

The sample 2 (control group), of size n2=100000 has a proportion of p2=0.0003.

p_2=X_2/n_2=30/100000=0.0003

The difference between proportions is (p1-p2)=-0.0001.

p_d=p_1-p_2=0.0002-0.0003=-0.0001

The pooled proportion, needed to calculate the standard error, is:

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The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.00026*0.99974}{100000}+\dfrac{0.00026*0.99974}{100000}}\\\\\\s_{p1-p2}=\sqrt{0+0.0000000025}=\sqrt{0.0000000051}=0.000071

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z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.00009-0}{0.000071}=\dfrac{-0.00009}{0.000071}=-1.26

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

P-value=P(z

As the P-value (0.104) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is  not  enough evidence to support the claim that the vaccine was effective.

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tankabanditka [31]

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Read 2 more answers
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This problem is a combination of the Poisson distribution and binomial distribution.

First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
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=6.18101*10^(-5)   or approximately
=0.00006181
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4 years ago
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