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KATRIN_1 [288]
3 years ago
6

azul made 5 veggie pizzas for his birthday party. he ordered enough for everyone to eat 5/6 of a pizza.how many people can he fe

ed?
Mathematics
2 answers:
sweet [91]3 years ago
6 0

Answer:

6 people

Explanation:

We divide the number of pizzas by the number of people to determine how much each person gets.

This means we are dividing 5 by an unknown number, x, and getting the answer 5/6.

This means the number of people, x, must be 6.

qaws [65]3 years ago
4 0
He can feed 6 PEOPLE
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9

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2 2/5

Step-by-step explanation:

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  9/8

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3 years ago
Ramon read twice on Saturday. In the morning, he read for
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if the numbers are separated by a fraction bar the response should be 11/12th of an hour on Saturday.

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4 0
3 years ago
Many states assess the skills of their students in various grades. One program that is available for this purpose is the Nationa
Nataly [62]

Answer:

A score of 314 is needed to be in the top 25% of students who take this exam.

Step-by-step explanation:

We are given that one of the tests provided by the NAEP assesses the reading skills of twelfth-grade students. In recent years, the national mean score was 289 and the standard deviation was 37.

Let X = <u><em>scores of the tests provided by the NAEP</em></u>

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = national mean score = 289

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Now, we have to find how high a score is needed to be in the top 25% of students who take this exam, that means;

            P(X > x) = 0.25     {where x is the required score}

            P( \frac{X-\mu}{\sigma} > \frac{x-289}{37} ) = 0.25

            P(Z > \frac{x-289}{37} ) = 0.25

In the z table, the critial value of z that represents the top 25% of the area is given as 0.6745, that is;

                        \frac{x-289}{37}=0.6745

                        {x-289}=0.6745\times 37

                        x = 289 + 24.96 = 313.96 ≈ 314

Hence, a score of 314 is needed to be in the top 25% of students who take this exam.

7 0
3 years ago
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