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marysya [2.9K]
3 years ago
8

Find the difference of 6/7 minus 1/14

Mathematics
2 answers:
VMariaS [17]3 years ago
5 0

Answer:

11/14

Step-by-step explanation:

TiliK225 [7]3 years ago
3 0

Answer:

11/14

Step-by-step explanation:

Convert to common denominator then subtract the numerator

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at Mr.neelys farm, there were 75 sheep and 60 cows what is the ratio number of cows to the number of sheep's at Mr.neelys farm
Charra [1.4K]

Answer:

The ratio of the number of cows to the number of sheeps in the farm is;

\begin{gathered} \frac{4}{5} \\ Or \\ 4\colon5 \end{gathered}

Explanation:

Given that;

there were 75 sheep

and 60 cows

We want to find the ratio of the number of cows to the number of sheeps in the farm;

\text{ratio}=\frac{\text{ number of cows }}{\text{ number of sh}eeps}=\frac{60}{75}

reducing the ratio to the least form;

\text{ratio=}\frac{60}{75}=\frac{4}{5}

Therefore, the ratio of the number of cows to the number of sheeps in the farm is;

\begin{gathered} \frac{4}{5} \\ Or \\ 4\colon5 \end{gathered}

3 0
1 year ago
A car in rush hour traffic travels 1/2 of a mile in 1/8 of an hour. If the car continues to travel at the same rate, how many mi
Maslowich
1/2 mile every 1/8 hour
Multiply by 8.
8/2 miles in 8/8 hours
4 miles per hour.
4 0
3 years ago
What is the correct integer for the following: a deposit of $1000?
Sladkaya [172]

Answer: One hundred thousand

 

Step-by-step explanation:

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6 0
3 years ago
Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.9 and a mean diameter of 200
ExtremeBDS [4]

Answer:

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 200, \sigma = 1.9, n = 78, s = \frac{1.9}{\sqrt{78}} = 0.2151

What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?

This is the pvalue of Z when X = 200 + 0.2 = 200.2 subtracted by the pvalue of Z when X = 200 - 0.2 = 199.8. So

X = 200.2

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{200.2 - 200}{0.2151}

Z = 0.93

Z = 0.93 has a pvalue of 0.8238

X = 199.8

Z = \frac{X - \mu}{s}

Z = \frac{199.8 - 200}{0.2151}

Z = -0.93

Z = -0.93 has a pvalue of 0.1762

0.8238 - 0.1762 = 0.6476

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

4 0
3 years ago
What is the area of the triangle?
alexandr402 [8]

Answer:

15/2

Step-by-step explanation:

Hello,

the area of a triangle is b*h/2

we can set BC as the base because AC is harder to calculate.

Then we draw an imaginary height.

There's no way to really explain, but the height would be 5 here, since 3 - -2 = 5

The base is 3

3*5/2 = 15/2

Thanks,

lacampanella

3 0
3 years ago
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