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Brut [27]
3 years ago
12

How could the relationship of the data be classified?

Mathematics
2 answers:
Mila [183]3 years ago
5 0

Answer: No correlation

Step-by-step explanation:

When we analyse any data and found that the two variables are not interdependent or proportional i.e. if one increasing then there is no proportional change in the other variable, then the relationship of the data  is classified as no correlation.

In the given picture, there is no proportional change in y with respect to x.

Therefore , the  relationship of the data is classified as " No correlation".

blagie [28]3 years ago
3 0
No correlation I believe.
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Liz got home from school at 4:33. She decided to bake muffins as an after-school snack. It took Liz 55 minutes to prepare and ba
IgorC [24]

Answer:

The muffins were done at 5:28

Step-by-step explanation:

you add 4:33 to 55 minutes knowing well that 60mins makes one hour

4:33

55

5:28

6 0
3 years ago
A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet
brilliants [131]

Answer:

There is not enough evidence to support the claim that the liquid diet yields a higher mean weight loss than the powder diet (P-value = 0.15).

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the liquid diet yields a higher mean weight loss than the powder diet.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0

The significance level is 0.05.

The sample 1 (powder diet group), of size n1=49 has a mean of 42 and a standard deviation of 12.

The sample 2 (liquid diet group), of size n2=36 has a mean of 45 and a standard deviation of 14.

The difference between sample means is Md=-3.

M_d=M_1-M_2=42-45=-3

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

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12^2}{49}+\dfrac{14^2}{36}}\\\\\\s_{M_d}=\sqrt{2.939+5.444}=\sqrt{8.383}=2.895

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-3-0}{2.895}=\dfrac{-3}{2.895}=-1.04

The degrees of freedom for this test are:

df=n_1+n_2-1=49+36-2=83

This test is a left-tailed test, with 83 degrees of freedom and t=-1.04, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.15) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the liquid diet yields a higher mean weight loss than the powder diet.

6 0
3 years ago
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