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3 years ago

Make an estimate. Write a number sentence for you estimate 927 divided by 8

Mathematics

1 answer:

notka56 [123]3 years ago

5 0

Answer:

estimated answer: 115

Step-by-step explanation:

By number sentence im guessing a word problem so if i'm wrong I apologize.

Johnny had 927 pokemon cards he wanted to get rid of so he shared them equally amongst 8 friends. If each friend received the same amount of cards than each friend shall receive about 115 cards

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lorasvet [3.4K]

Answer:

Csc(pi)

Cot(pi)

Csc(0)

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Step-by-step explanation:

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3 years ago

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Georgia [21]

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Step-by-step explanation:

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3 years ago

Can someone please help me with 44. And 45?

Angelina_Jolie [31]

Problem 44

The term "bisect" means "cut in half".

Since BD bisects angle ABC, this means the smaller angles ABD and DBC are congruent.

angle ABD = angle DBC

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<h3>Answer: x = 20</h3>

========================================================

Problem 45

We use the same idea as the previous problem

angle ABD = angle DBC

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2 years ago

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Oksanka [162]

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Step-by-step explanation:

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3 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:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

3 years ago