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Vaselesa [24]
3 years ago
9

What is the spread of the data?

Mathematics
1 answer:
Nonamiya [84]3 years ago
5 0
C is the correct answer
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The table shows a proportional relationship between the number of meals
Kisachek [45]
B is the answer there is a common denominator y’a welcome
4 0
3 years ago
Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
Alborosie

Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

6 0
3 years ago
two numbered 1 to 6 dice are thrown together and their scores are added. the probability that the sum will be 4 is?​
natulia [17]

Answer:

Pr = \frac{3}{36}

Step-by-step explanation:

Given

2 number die

Required

P(Sum = 4)

The sample space of 2 die is:

S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6), (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

n(S)= 36

The pairs that adds up to 4 are:

Sum(4) = \{(1,3), (2,2),(3,1)\}

n(Sum)  =3

So, the probability is:

Pr = \frac{n(Sum)}{n(S)}

Pr = \frac{3}{36}

5 0
2 years ago
Please help me with this
jekas [21]

Answer:

The length= 16cm

The width= 6cm

The two long sides are 8cm both so 8 + 8 = 16

And the two short sides are 3cm both so 3 + 3 = 6

So your length should be 16cm and your width 6cm

I HOPE THIS HELPS! ∧   ∧

                              ⊂∵→ω←∵⊃

3 0
3 years ago
WILL MARK BRAINLEST HELPP ME
vitfil [10]

Answer:

Supplimentry

Step-by-step explanation:

7 0
3 years ago
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