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Deffense [45]
2 years ago
15

Peter wants to purchase pizza pies and breadsticks for a party. The cashier tells him that pizza pies are $8 each and breadstick

s are $5 each. Peter cannot spend more than $120. Which of the following options models the number of pizza pies and breadsticks that Peter can purchase? Let y represent the number of pizza pies purchased and let x represent the number of breadsticks purchased. 5x + 8y ≤ 120 5x + 8y ≥ 120 8x + 5y ≤ 120 8x + 5y ≥ 120
Mathematics
2 answers:
kolezko [41]2 years ago
7 0
Apromixly this would equal 5x+8y<span>≤120</span>
sladkih [1.3K]2 years ago
6 0
The answer is 5x+8y<= 120
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Jenna wants to open a salon and must choose a location. At Location A, she will pay $1200 per month and charge $45 per haircut.
anzhelika [568]
Profit=revenue-cost

pA(m,h)=45h-1200m

pB(m,h)=60h-1800m

Profit is the same when pA=pB so

60h-1800m=45h-1200m

15h=600m

h=40m

So 40 haircuts per month at either location results in the same profit of $600 per month.
5 0
2 years ago
Juliet rented a car for one day from a company that charges $80 per day plus $0.15 per mile driven. If she was charged a total o
svlad2 [7]

Answer:

B) 120

Step-by-step explanation:

m = miles

$98=$80+$0.15m

$98-$80=$80-$80+$0.15m

$18=$0.15m

$18/$0.15=$0.15m/$0.15

120=m

Juliet drove 120 miles for $98.

(this is a good explanation, I guess)

7 0
2 years ago
Read 2 more answers
The probabilities are 0.4, 0.2, 0.3 and 0.1, respectively, that a delegate to a certain convention arrived by air, bus, automobi
Anika [276]

Answer:

See solution solved accordingly

Step-by-step explanation:

5 0
2 years ago
Kurtosis of a normal data distribution is a ___________________ Group of answer choices Measure of data centrality Measure of da
Irina18 [472]

Answer:

Option A, Measure of data centrality

Step-by-step explanation:

Normal distribution curve determines the skewness of the data as to what extent is is disoriented from the normal distribution curve. Also, it is the probability distribution curve that determines whether the data set is two legged, one legged or it lies under which probability curve.

Hence option A is correct

5 0
2 years ago
A random sample of the actual weight of 5-lb bags of mulch produces a mean of 4.963 lb and a standard deviation of 0.067 lb. If
dsp73

Answer: B) 4.963±0.019.

Step-by-step explanation:

Confidence interval for population mean ( when population standard deviation is not given) is given by :-

\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}

, where \overline{x} = Sample  mean

n= Sample size

s= sample standard deviation

t* = critical t-value.

As per given:

n= 50

Degree of freedom = n-1 =49

\overline{x}= 4.963\ lb

s= 0.067 lb

For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.

Now , substitute all values in the formula , we get

4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019

Hence,  a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is 4.963\pm0.019.

Thus , the correct answer is B) 4.963±0.019.

7 0
2 years ago
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