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3 years ago

Write a ratio to compare the amounts below. Simplify

Mathematics

2 answers:

Taya2010 [7]3 years ago

8 0

Answer:

187/1200

Step-by-step explanation:

9.35 minutes = 561 s

1 hr = 3600 s

561 / 3600 = 187/1200

frutty [35]3 years ago

3 0

Answer:

6/25

Step-by-step explanation:

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she had 79.2 dollars before she purchased the songs. Each song costs -24.

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2 years ago

Samantha is using a 2-liter pitcher to serve lemonade to 10 of her friends. How many times will she need to fill the pitcher in

Mice21 [21]

Answer: 2 times

Step-by-step explanation:

Hi, since 1000 milliliters = 1 liter

We have to convert the 2 liters into milliliters:

2 x 1000 = 2000 milliliters, so she is using a 2,000 milliliters pincher.

Now we have to multiply the number of friends (10) by the amount of lemonade she serves to each one (400 ml)

400 x 10 = 4,000 milliliters

Finally, we have to divide the amount of lemonade she needs to serve (4,000 ml) by the capacity of the pitcher (2,000), to obtain the number of times that she will need to fill the pitcher.

4,000/2,000=2 times

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3 years ago

Read 2 more answers

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95