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4 years ago

Find the margin of error for each sample. Then find an interval likely to contain the true population proportion.

Mathematics

1 answer:

adelina 88 [10]4 years ago

4 0

A.15% of 475 teacher is the answer

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A used car is priced at $5250. If the price is dropped to $4500, what is the percent of change in the price of the car? Round yo

saw5 [17]

85.7%
4500/5250=.857
move decimal 2 places to the right to convert a decimal to a percentage

8 0

3 years ago

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Select the term that is suggested by the figure

bulgar [2K]

I think it’s line segment

5 0

2 years ago

I need help I dont know how to do this

Nikolay [14]

Answer:

9.2

Step-by-step explanation:

you get 9.1651513... but it says round to the nearest 10th which would be 9.2

7 0

3 years ago

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4400 dollars is placed in an account with an annual interest rate of 7.5%. To the nearest tenth of a year, how long will it take

Katarina [22]

Answer:

Step-by-step explanation:

The equation for the amount of money in an account after a certain amount is deposited and compounded after t years once per year is

$A(t)=P(1+r)^t$

Our A(t) = 33800, P = 4400, r = .075 and we are looking for t. Filling in:

$33800=4400(1+.075)^t$ and

$33800=4400(1.075)^t$

Begin by dividing both sides by 4400 to get

$7.681818182=1.075^t$

The only way to move that t our from its current position as an exponent is to take the natural log of both sides and follow the rules for natural logs:

$ln(7.6181818182=ln(1.075)^t$

The power rule of natural logs says we can move that exponent down in front, giving us:

$ln(7.681818182)=t*ln(1.075)$

Divide both sides by ln(1.075) to get

$\frac{ln(7.61818182)}{ln(1.075)} =t$

Do this division on your calculator to get

t = 28.2 years

6 0

3 years ago

Read 2 more answers

A Boxer. needs to lose 3 1/2 kg in a month to be able to complete as a flyweight. In three weeks he lowers his weight from 55.5

Airida [17]

Answer: The boxer lose 1.8 kg in the final week to get able to compete as a flyweight.

Step-by-step explanation:

Since we have given that

Quantity of weight a boxer needs to lose in a month is given by

$3\frac{1}{2}\ kg\\\\=\frac{7}{2}\ kg=3.5\ kg$

In 3 months, he lowers his weight from 55.5 kg to 53.8 kg.

So, Quantity of weight he lose in three months is given by

$55.5-53.8\\\\=1.7\ kg$

Number of kilograms the boxer lose in the final week to be able to compete as a flyweight is given by

$3.5-1.7\\\\=1.8\ kg$

Hence, the boxer lose 1.8 kg in the final week to get able to compete as a flyweight.

4 0

3 years ago

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