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EastWind [94]
3 years ago
8

Nola purchased 2.5 pounds of cheese for $10.50. Her mother purchased 3 pounds of the same cheese for $12.60. The cost of cheese

varies directly with the number of pounds purchased . How much does one pound of cheese cost?
Mathematics
1 answer:
jeka943 years ago
4 0

Answer:

cost per pound = $4.20

Step-by-step explanation:

Cheese cost per pound = ?

first for the cost per pound

cost per pound = total cost / pounds bought

cost per pound = $10.50 / 2.5 pounds

cost per pound = $4.20

to check the above answer is correct

cost per pound = total cost / pounds bought

cost per pound = $12.60 / 3 pounds

cost per pound = $4.20

Therefor, the cheese costs $4.20 per pound.

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Write two equivalent equations for the equation 3f+2m=16.
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3 years ago
A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. ​a) E
polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02

So, replacing the values we get:

0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

It means the the level of significance \alpha is:

\alpha =1-0.98=0.02

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