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stira [4]
2 years 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]2 years 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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$4.081

Step-by-step explanation:

We have been given that Melissa owns a credit card with a 8.9% APR. The balance after her last billing cycle was $550.      

First of all, we will find period rate by dividing APR by 12 as 1 years equals 12 months.

\text{Period rate}=\frac{8.9\%}{12}=0.74166\%

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CyberCafe charges a computer station rental fee of $5 plus $0.20 for each quarter hour spent surfing. Which expression represent
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Searchers are planning a study to estimate the impact on crop yield when no-till is used in combination with residue retention a
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Answer:

n=77

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Notation and definitions

\hat p=0.025 or 2.5%, estimated proportion for decline in crop yield when no-till is used with residue retention and crop rotation

p true population proportion for decline in crop yield when no-till is used with residue retention and crop rotation

n (variable of interest) represent the sample size required

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.035 or 3.5% and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

And replacing into equation (b) the values from part a we got:

n=\frac{0.025(1-0.025)}{(\frac{0.035}{1.96})^2}=76.44  

And rounded up we have that n=77

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