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3 years ago

HELP! GIVING BRAINLIEST AND POINTS!!!

1 answer:

5 0

1. 8 1/3 is greater.
Least to Greatest it would be 7 1/5, 7 1/2, and 7 7/8. This is because 1/5 is less than 1/2, and 1/2 is less than 7/8.

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Jenna jump 156 inches in the long jump competition After High School track meet how many feet did Jenna jump

klio [65]

13 feet hears how:
156/12=13

divide it

Hope this helps

7 0

2 years ago

Read 2 more answers

Identify the triangles that are similar by the AA similarity theorem. explain how you know that the triangles are similar.

UNO [17]

Angle WVX = Angle YVZ (vertically opposite angles)

Angle VWX = Angle VYZ (shown)

Angle WXV = Angle YZV (total sum of angles in triangle)

4 0

2 years ago

Solve the absolute value equation |8|=28

DENIUS [597]

The answer is false. This is because 8 does not equal the number 28.

Explination:

|8| = 28

8≠28

7 0

2 years ago

How do i write 7.6% as a fraction in simplest form and as a decimal?

egoroff_w [7]

You would move the decimal point two times to the left for the decimal. 0.076
For the fraction, we would make it 76/1000 because the 6 converted in the decimal is in the thousandth place. We can simplify it and get 19/250.

6 0

2 years ago

Read 2 more answers

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)

skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Step-by-step explanation:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample x⁻ = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

Step 2:-

The 95% of confidence intervals for the average spending

( $(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

$(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )$

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{28-30 }{\frac{10}{\sqrt{25} } }$

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

Conclusion:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0

2 years ago