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Murljashka [212]
3 years ago
8

Write each unit rate as a fraction in lowest terms 15miles in 10 hours

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

im pretty sure its 1.5


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Find the value of x. x√5=x^4​
Kaylis [27]

Answer:

x=0,\sqrt[6]{5}

Decimal form,

x=0,1.30766048......

Step-by-step explanation:

x\sqrt{5}=x^{4}

x\sqrt{5}-x^{4}=0

(take x as a common factor)

x(\sqrt{5}-x^{3})=0

So,

x=0 and \sqrt{5}-x^{3}=0

for \sqrt{5-x^{3}=0

(add x^{3} for both sides)

\sqrt{5}=x^{3}

x=\sqrt[3]{\sqrt[2]{5} } (multiply 3 by 2)

x=\sqrt[6]{5}

So,

x=0,\sqrt[6]{5}

5 0
3 years ago
What is 14.62 to the nearest whole number
alexandr1967 [171]
Because the decimal is over 40 the answer is 15
6 0
3 years ago
Read 2 more answers
Solve the system of equations. Write you answer as an ordered pair.
lawyer [7]

Answer:

x = -4    y = -5

Step-by-step explanation:

2x + 3 = -x - 9

3x + 3 = -9

3x = -12

x = -4

y = - (-4) - 9

y = 4 - 9

y = -5

6 0
3 years ago
PLEASE HELP WILL MARK BRAINLIEST
Romashka [77]

Answer:

A.The mean would increase.

Step-by-step explanation:

Outliers are numerical values in a data set that are very different from the other values. These values are either too large or too small compared to the others.

Presence of outliers effect the measures of central tendency.

The measures of central tendency are mean, median and mode.

The mean of a data set is a a single numerical value that describes the data set. The median is a numerical values that is the mid-value of the data set. The mode of a data set is the value with the highest frequency.

Effect of outliers on mean, median and mode:

  • Mean: If the outlier is a very large value then the mean of the data increases and if it is a small value then the mean decreases.
  • Median: The presence of outliers in a data set has a very mild effect on the median of the data.
  • Mode: The presence of outliers does not have any effect on the mode.

The mean of the test scores without the outlier is:

   \bar{x}=\frac{Total of the observations-Outlier value}{n-1} \\=\frac{(86*16)-72}{15} \\=\frac{1304}{15}\\ =86.9333

*Here <em>n</em> is the number of observations.

So, with the outlier the mean is 86 and without the outlier the mean is 86.9333.

The mean increased.

Since the median cannot be computed without the actual data, no conclusion can be drawn about the median.

Conclusion:

After removing the outlier value of 72 the mean of the test scores increased from 86 to 86.9333.

Thus, the the truer statement will be that when the outlier is removed the mean of the data set increases.

4 0
3 years ago
Please help <br> Thank you
qwelly [4]

Answer:

A series of prime numbers starting with 9?

8 0
3 years ago
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