An orthocenter of a triangle is

Mathematics

1 answer:

Kamila [148]2 years ago

5 0

Answer:

the intersection of the triangle's three altitudes.

Step-by-step explanation:

hope this helped </3

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A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The owner of the store has d

Alinara [238K]

Answer:

Read step by step explanation

Step-by-step explanation:

The owner already knows that the limit for the average time delivered pizzas is 38 minutes. So we conclude

1.-The resulting mean from sample data ( x ) ( 27 customers) need to be smaller than 38 minutes, any value of sample above 38 minutes means more time for the delivery action and will indicate a failure for the future project

2.-As sample size is smaller than 30 the test has to be t-student one tail test to the left

Test hypothesis

Null hypothesis H₀ x = 38

Alternative hypothesis Hₐ x < 38

We should test at a significance level α = 0,05 (α = 5%)

If the result of the test is to accept H₀ delivery project won´t be implemented, if on the other hand, H₀ is rejected then in the condition of the alternative hypothesis we accept Hₐ the sample indicates that we have a smaller average time than 38 minutes.

4 0

2 years ago

I need the answer for this problem please

Savatey [412]

Answer:

Step-by-step explanation:

Find all probabilities:

A. False

$Pr(\text{red shirt}|\text{large shirt})=\dfrac{\text{number red large shirts}}{\text{number large shirts}}=\dfrac{42}{77}=\dfrac{6}{11}\\ \\Pr(\text{large shirt})=\dfrac{\text{number large shirts}}{\text{number shirts}}=\dfrac{77}{165}=\dfrac{7}{15}$

B. True

$Pr(\text{blue shirt}|\text{large shirt})=\dfrac{\text{number blue large shirts}}{\text{number large shirts}}=\dfrac{35}{77}=\dfrac{5}{11}\\ \\Pr(\text{blue shirt})=\dfrac{\text{number blue shirts}}{\text{number shirts}}=\dfrac{75}{165}=\dfrac{5}{11}$

C. False

$Pr(\text{shirt is medium and blue})=\dfrac{\text{number medium and blue shirts}}{\text{number shirts}}=\dfrac{48}{165}=\dfrac{16}{55}\\ \\Pr(\text{medium shirt})=\dfrac{\text{number medium shirts}}{\text{number shirts}}=\dfrac{88}{165}=\dfrac{8}{15}$

D. False

$Pr(\text{large shirt}|\text{red shirt})=\dfrac{\text{number red large shirts}}{\text{number red shirts}}=\dfrac{42}{90}=\dfrac{7}{15}\\ \\Pr(\text{red shirt})=\dfrac{\text{number red shirts}}{\text{number shirts}}=\dfrac{90}{165}=\dfrac{6}{11}$