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3 years ago

572000 square miles rounded ro the nearest hundred thousand

1 answer:

7 0

It would be rounding it to 600000 because 600000 is the closest 100000x amount where x is an integer. I hope this helped and if I am right could I have brainliest? :)

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June bought three books. For the first book, she paid one half of her allowance plus $1 more. For the second book, she paid one

inysia [295]

Answer:

$34

Step-by-step explanation:

I'd work backwards.

For the third book, she paid all her remaining money. The problem says she paid "1/2 her leftover money + $3". This means that: (let m = money used to buy book 3)

m = 1/2m + 3

1/2m = 3

m = 6

For the second book: (let n = money before book 2)

n - m (money left after book 2) = 1/2n + 2

1/2n +2 is money used up for book, which is the same as n-m.

n = 1/2n + 2 + m

1/2n = 2 + m = 2 + 6

1/2n =8

n = 16

For the first book, she spent 1/2 her money + 1. If o = money before book 1 (or the whole allowance):

o - n = 1/2o + 1

o = 1/2o + n + 1 = 1/2o + 16 + 1

1/2o = 17

o = 34

Check!

Spent 17 (half 34) + 1 on book 1

16 left

Spent 8 (half 16) +2 on book 2

6 left

Spent 3 (half 6) + 3 on book 3

0 left

4 0

3 years ago

According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

Let $\hat p$ = sample proportion of Americans who can order a meal in a foreign language

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }$ ) = P(Z > 0.85) = 1 - P(Z $\leq$ 0.85)

= 1 - 0.80234 = 0.19766

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0

3 years ago

What does 3 exponent -2 equal

inysia [295]

I am pretty sure your answer is going to be 243
I hope this helps, and good luck!!

4 0

3 years ago

How long does it take to drive a distance of 474 miles at a speed of 75 mph?

lisabon 2012 [21]

Answer:

6.32 hours