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stira [4]
1 year ago
5

Jacque needs to buy some pizzas for a party at her office. She's ordering from a restaurant that charges a \$7.50$7.50dollar sig

n, 7, point, 50 delivery fee and \$14$14dollar sign, 14 per pizza. She wants to buy as many pizzas as she can, and she also needs to keep the delivery fee plus the cost of the pizzas under \$60$60dollar sign, 60.
Each pizza is cut into 888 slices, and she wonders how many total slices she can afford.

Let PPP represent the number of pizzas that Jacque buys.

1) Which inequality describes this scenario?
Mathematics
1 answer:
BigorU [14]1 year ago
5 0

Answer:

Step-by-step explanation:

Delivery fee = $7.50

Price per pizza = $14

Planned amount = $60

7.50 + 14p < 60

Subtract 7.50 from both sides

14p < 60 - 7.50

14p < 52.50

Divide both sides by 14

p < 52.50 / 14

p < 3.75

Each pizza is cut into 8 slices

Total slices she can afford = 3.75 × 8

= 30 slices

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A recipe calls for 1/4 cup of broth. The largest volume-measuring tool that Tabitha has is a teaspoon. She knows that there are
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Option b. 12 teaspoons.

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In the question it given :

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According to a Pew Research survey, about 27% of American adults are pessimistic about the future of marriage and the family. Th
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Answer:

P(X≤5)=0.5357

Step-by-step explanation:

Using the binomial model, the probability that x adults from the sample, are pessimistic about the future is calculated as:

P(x)=\frac{n!}{x!(n-x)!} *p^{x}*(1-p)^{n-x}

Where n is the size of the sample and p is the probability that an adult is pessimistic about the future of marriage and family. So, replacing n by 20 and p by 0.27, we get:

P(x)=\frac{20!}{x!(20-x)!}*0.27^{x}*(1-0.27)^{20-x}

Now, 25% of 20 people is equal to 5 people, so the probability that, in a sample of 20 American adults, 25% or fewer of the people are pessimistic about the future of marriage and family is equal to calculated the probability that in the sample of 20 adults, 5 people of fewer are pessimistic about the future of marriage and family.

Then, that probability is calculated as:

P(X≤5)= P(1) + P(2) + P(3) + P(4) + P(5)

Where:

P(0)=\frac{20!}{0!(20-0)!}*0.27^{0}*(1-0.27)^{20-0}=0.0018

P(1)=\frac{20!}{1!(20-1)!}*0.27^{1}*(1-0.27)^{20-1}=0.0137

P(2)=\frac{20!}{2!(20-2)!}*0.27^{2}*(1-0.27)^{20-2}=0.0480\\P(3)=\frac{20!}{3!(20-3)!}*0.27^{3}*(1-0.27)^{20-3}=0.1065\\P(4)=\frac{20!}{4!(20-4)!}*0.27^{4}*(1-0.27)^{20-4}=0.1675\\P(5)=\frac{20!}{5!(20-5)!}*0.27^{5}*(1-0.27)^{20-5}=0.1982

Finally, P(X≤5) is equal to:

P(X≤5) = 0.0018+0.0137 + 0.0480 + 0.1065 + 0.1675 + 0.1982

P(X≤5) = 0.5357

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