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2 years ago

10

Pedro owns a shrimp truck near the beach. He sells garlic shrimp for $8 a plate and spicy shrimp for $6 a plate. Each day Pedro

stocks enough shrimp to sell at most 120 plates total, but he would like to earn at least $800. Which combination of garlic shrimp plates and spicy shrimp plates can Pedro sell to meet his goal? A. 15 garlic shrimp plates and 110 spicy shrimp plates B. 80 garlic shrimp plates and 20 spicy shrimp plates C. 90 garlic shrimp plates and 25 spicy shrimp plates D. 70 garlic shrimp plates and 55 spicy shrimp plates

1 answer:

yKpoI14uk [10]2 years ago

7 0

Answer:

C

Step-by-step explanation:

You can plug in each combination to see what would work

C is the only combination that satisfies all the constraints: it earns him more than $800 and is less than 120 plates.

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(b) A department store has 7,000 charge accounts. The comptroller takes a random sample of 36 of

galben [10]

Answer:

a.0.8664

b. 0.23753

c. 0.15866

Step-by-step explanation:

The comptroller takes a random sample of 36 of the account balances and calculates the standard deviation to be N42.00. If the actual mean (1) of the account balances is N175.00, what is the probability that the sample mean would be between

a. N164.50 and N185.50?

b. greater than N180.00?

c. less than N168.00?

We solve the above question using z score formula

z = (x-μ)/σ/√n where

x is the raw score,

μ is the population mean = N175

σ is the population standard deviation = N42

n is random number of sample = 36

a. Between N164.50 and N185.50?

For x = N 164.50

z = 164.50 - 175/42 /√36

z = -1.5

Probability value from Z-Table:

P(x = 164.50) = 0.066807

For x = N185.50

z = 185.50 - 175/42 /√36

z =1.5

Probability value from Z-Table:

P(x=185.50) = 0.93319

Hence:

P(x = 185.50) - P(x =164.50)

= 0.93319 - 0.066807

= 0.866383

Approximately = 0.8664

b. greater than N180.00?

x > N 180

Hence:

z = 180 - 175/42 /√36

z = 5/42/6

z = 5/7

= 0.71429

Probability value from Z-Table:

P(x<180) = 0.76247

P(x>180) = 1 - P(x<180) = 0.23753

c. less than N168.00?

x < N168.

z = 168 - 175/42 /√36

z = -7/42/6

z = -7/7

z = -1

Probability value from Z-Table:

P(x<168) = 0.15866

4 0

1 year ago

Say a certain service industry has 78.9 thousand jobs in 2003, but expects to increase at an average annual rate of 2.65 thousan

Alex_Xolod [135]

Answer:

33%

Step-by-step explanation:

In 2003, there were 78.9 thousand jobs in the industry and it increases annually by 2.65 thousand jobs from 2003 to 2013.

Hence, from 2003 to 2013, i.e. 10 years the jobs have increased by (2.65×10) = 26.5 thousand.

Therefore, the percent increase in jobs in the industry from 2003 to 2013 is given by $\frac{26.5*100}{78.9} = 33.6$ % ≈ 33%. (Answer)

4 0

1 year ago

Jenna works in an ice cream shop. When she starts her shift the tub of chocolate ice cream is 23 full. When she finishes her shi

ipn [44]

Answer:

<em /> $Rate = 1.1/hr$ <em />

<em />

Step-by-step explanation:

Given

$Start = 23$

$End = 16$

$Hours = 6\frac{1}{2}h$

Required

Determine the hourly rate

First, we need to determine the ice cream sold

$Ice\ Cream = 23 - 16$

$Ice\ Cream = 7$

Next, we calculate the hourly rate:

$Rate = Ice\ Cream/Hours$

$Rate = 7/6\frac{1}{2}h$

$Rate = 7/\frac{13}{2}h$

$Rate = 7*\frac{2}{13h}$

$Rate = \frac{14}{13h}$

$Rate = 1.07692307692/hr$

<em /> $Rate = 1.1/hr$ <em> -- Approximated</em>

4 0

1 year ago

Read 2 more answers

production for five people was as follows: 8 units, 11 units, 6 units, 12 units, 8 units. what was their average production in u

zlopas [31]

Add all the units together 8+11+6+8=45
then divide by the the number of units there are which is 5 so 45÷5 =9

4 0

2 years ago

Read 2 more answers

In measuring reaction time, a psychologist estimates that a standard deviation is .05 seconds. How large a sample of measurement

blondinia [14]

Answer:

97

Step-by-step explanation:

We are asked to find the size of sample to be 95% confident that the error in psychologist estimate of mean reaction time will not exceed 0.01 seconds.

We will use following formula to solve our given problem.

$n\geq (\frac{z_{\alpha/2}\cdot\sigma}{E})^2$ , where,

$\sigma=\text{Standard deviation}=0.05$ ,

$\alpha=\text{Significance level}=1-0.95=0.05$ ,

$z_{\alpha/2}=\text{Critical value}=z_{0.025}=1.96$ .

$E=\text{Margin of error}$

$n=\text{Sample size}$

Substitute given values:

$n\geq (\frac{z_{0.025}\cdot\sigma}{E})^2$

$n\geq (\frac{1.96\cdot0.05}{0.01})^2$

$n\geq (\frac{0.098}{0.01})^2$

$n\geq (9.8)^2$

$n\geq 96.04$

Therefore, the sample size must be 97 in order to be 95% confident that the error in his estimate of mean reaction time will not exceed 0.01 seconds.

5 0

2 years ago