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FrozenT [24]
3 years ago
6

Lenz father is paying for a $20 meal. He has a 15%-off coupon on his meal. After the discount, a 7% sales tax is applied. How mu

ch does Lynn’s father pay for the meal
Mathematics
1 answer:
-BARSIC- [3]3 years ago
3 0

Answer:

18.19.

First you have to subtract 15 from 100, giving you 85%. You then multiple 85% to the $20, giving you $17. You then add the 7 to 100, giving you 1.07%, then you multiple that to the $17 and you end up with $18.19.

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Solve and simplify.. 1/2 ÷6 =​
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Answer:

1/12

Step-by-step explanation:

1/2 ÷ 6

is the same as;

1/2 × 1/6 = 1/12

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Erin wants to measure the length of a pond. She measured 15 yd from point Y to point Z and 17 yd from point X to point Z. What i
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X the length of the bounce is 25
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Four brothers and their sister have the average age 7. Four brothers have the average age 6. How old is the sister?
AnnyKZ [126]

Answer:

11 years old

Step-by-step explanation:

If four brothers have the average age 6, then the sum of their ages is

4\cdot 6=24

years.

Let x years be the age of sister. If four brothers and their sister have the average age 7, then

\dfrac{24+x}{5}=7.

Multiply this equation by 5:

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The sister is 11 years old.

4 0
3 years ago
What is 2/3 to the power of 3?
VikaD [51]

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Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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

4 0
3 years ago
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