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strojnjashka [21]
3 years ago
10

Which table shows a proportional relationship between x and y? (a)x y 1 5 2 6 3 7 (b)x y 2 3 4 6 6 9 (c) x y 4 16 8 64 12 144 (d

)x y 1 3 2 9 3 27
Mathematics
1 answer:
JulijaS [17]3 years ago
7 0
A) 0 , 4
B) 0 , 0
C) 0 , 0
D) 0 , 1

Are you kidding
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Scorpion4ik [409]

Answer:

x = 7.2m

Step-by-step explanation:

Because it is a right triangle, you can use the Pythagorean theorem:

4^2+6^2 = x^2

16+36 = x^2

52 = x^2

sqrt52 = x

7.2 = x

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the band director is splitting 1\4 of the band into 4 equal rehearsal group what fraction of the entire band will be in each reh
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Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last
lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55

The degrees of freedom are df=599

df=n-1=600-1=599

The P-value for this test statistic is:

P-value=P(t

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

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I will give brainliest to who answers this. Please answer quick
Kazeer [188]

Answer:

Im pretty sure it would just be g= -p+15

Step-by-step explanation:

I really hope this helps, if not I am so sorry

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Look at attached screenshot!

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