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3 years ago

60 divided by 886 divided by 615 divided by 656 divided by 2

1 answer:

6 0

60÷8= 7 remainder 4
86÷6= 14 remainder 2
15÷6= 2 remainder 3
56÷2= 28

If you're diving by decimals then...

60÷8= 7.5
86÷6= 14.3 Draw a line over the number like this ___
14.3
15÷6=2.5
56÷2= 28.0 or 28.

Hope this helps!!

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Help answer this im struggling omg

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3 years ago

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What’s the reminder for 90 divide by 4

evablogger [386]

The answer is 22.5 so the reminder will be 1/2 or .5

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3 years ago

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Robert borrowed $750 from his parents. He promised to repay the loan by giving his parents at least $67 from his paycheck each w

Luda [366]

To answer this question we can propose the following equation:

$d=750 -x(67)$

Where:

x is the number of weeks elapsed.

d is the debt depending on the weeks.

Robert promised to pay at least $ 67 each week.

Therefore, after 5 weeks the debt of rober should be:

$d\leq750-5*67$

$d\leq415$

Then scenario A) is possible because $150$ .

After 6 weeks:

$d\leq750-6*67$

$d\leq348$

Then scenario B) is possible because $200$ .

After 7 weeks

$d\leq750-7*67$

$d\leq281$

Then scenario C) is possible because $250$ .

After 8 weeks

$d\leq750-8*67$

$d\leq214$

So scenario D) is NOT possible because $300>214$

Finally the correct answer is option D.

3 0

2 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: