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Talja [164]
3 years ago
12

Jasmine bought 1.3 pounds of ham and 0.8 pounds of cheese from the deli and paid $11.28. She went back the following week and bo

ught 1.5 pounds of ham and 1.2 pounds of cheese and paid $14.76. If the prices remained the same, find the price per pound of ham and cheese using the elimination method
Mathematics
1 answer:
LuckyWell [14K]3 years ago
5 0
Let

x
be the price for a pound of ham
y the price for a pound of cheese

<span>1.3x +0.8y = 11.28  ....(1)
1.5x +1.2y = 14.76  ....(2)

Lets multiply (1) by -1.2
Lets multiply (2) by 0.8

-1.56x -0.96y = -13.536 .....(3)
</span> 1.2x + 0.96y = 11.808 .....(4)


If we add (3) and (4)

-0.36x = -1.728 ............> x = 4.8

We substitute in (1)

1.3(4.8) +0.8y = 11.28

y = 6.3

Price per pound of ham = x = 4.8
Price per pound of cheese = y = 6.3
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The rate of change in temperature is −17°F/min

<h3>Rate of change of a function</h3>

The formula for calculating the rate of change of a function is expressed as:

Rate of change = rise/run

This is also known as the slope of the function

Given the following parameters

Temperature = −25.5°F

Cooling time = 1.5mins

Determine the rate of change

Rate = −25.5°F/1.5min

Rate = −17°F/min

Hence the rate of change in temperature is −17°F/min

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2 years ago
What id the greatest common factor shared by 24 and 54
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6.

Step-by-step explanation:

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3 years ago
A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. ​(a) Describe the
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Answer:

a) \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

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b) z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

c) z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

d) z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

\mu = 83, \sigma = 27

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

Part b

We want this probability:

P(\bar X>89)

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 89 we got:

z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

Part c

P(\bar X

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 75.65 we got:

z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

Part d

We want this probability:

P(79.4 < \bar X < 89.3)

We find the z scores:

z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

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Answer:

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Step-by-step explanation:

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The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. What percent of the scores are
Lynna [10]
A normal distribution with a mean of 74 and a standard deviation of 7.
Mean + 1 SD = 74 + 7 = 81
Less than 81 :  50 % + 34 % = 84 %
Answer:
A ) 84 % 
4 0
3 years ago
Read 2 more answers
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