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3 years ago

8

Tammy ran for president of the chess club and she received 27 votes they were 45 members in the club what percentage of the club

members voted for Tammy?

1 answer:

Sonbull [250]3 years ago

8 0

Answer:

60% of the club members voted for Tammy

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What is the area of the base. explain, please :)

Ber [7]

Answer:

4,200

Step-by-step explanation:

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3 years ago

A snack tray at a party has cheese squares with 2 grams of protein apiece and turkey slices with 3 grams of protein apiece. whic

Dafna11 [192]

The inequality that represents the possible ways nina can eat 12 or more grams of protein is 2x + 3y ≥ 12

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more number and variables.

Let x be the number of cheese squares that she eats and y is the number of turkey slices that she eats. If nina can eat 12 or more grams of protein, hence:

2x + 3y ≥ 12

The inequality that represents the possible ways nina can eat 12 or more grams of protein is 2x + 3y ≥ 12

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2 years ago

Pq + 4; use p=4, and q = 4

dangina [55]

Answer:

20

Step-by-step explanation:

pq + 4

Given:

p = 4

q = 4

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pq + 4

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20

5 0

3 years ago

Read 2 more answers

Enter an algebraic expression for the word expression.<br> twice a number x, minus 16

steposvetlana [31]

Answer:

x(2) - 16 or (2)x - 16

Step-by-step explanation:

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3 years ago

Find the following list of data calculate a demean be the maid in and see the mouth or mothers of the numbers from

Lelechka [254]

Given:

A fair die is rolled.

It pays off $10 for 6, $7 for a 5, $4 for a 4 and no payoff otherwise.

To find:

The expected winning for this game.

Solution:

If a die is rolled then the possible outcomes are 1, 2, 3, 4, 5, 6.

The probability of getting a 6 is:

$P(6)=\dfrac{1}{6}$

The probability of getting a 5 is:

$P(5)=\dfrac{1}{6}$

The probability of getting a 4 is:

$P(4)=\dfrac{1}{6}$

The probability of getting other numbers (1,2,3) is:

$P(\text{Otherwise})=\dfrac{3}{6}$

$P(\text{Otherwise})=\dfrac{1}{2}$

Now, find the sum of product of payoff and their corresponding probabilities to find the expected winning for this game.

$E(x)=10\times P(6)+7\times P(5)+4\times P(4)+0\times P(\text{Otherwise})$

$E(x)=10\times \dfrac{1}{6}+7\times \dfrac{1}{6}+4\times \dfrac{1}{6}+0\times \dfrac{1}{2}$

$E(x)=\dfrac{10}{6}+\dfrac{7}{6}+\dfrac{4}{6}+0$

On further simplification, we get

$E(x)=\dfrac{10+7+4}{6}$

$E(x)=\dfrac{21}{6}$

$E(x)=3.5$

Hence, the expected winnings for this game are $3.50.

5 0

3 years ago