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goldenfox [79]
3 years ago
8

Bill reads 1/5 of a book on Monday he reads 2/3 of a book on Tuesday if he finish reading the book on Wednesday what fraction of

the book did he read on Wednesday
Mathematics
1 answer:
Evgen [1.6K]3 years ago
5 0

total book =1

1/5 + 2/3 + wed = 1

get a common denominator of 15

1/5 * 3/3 + 2/3 * 5/5 + wed = 1 * 15/15

3/15 + 10/15 + wed = 15/15

combine like terms

13/15 + wed = 15/15

subtract 13/15 on each side

wed = 15/15-13/15

wed = 2/15

He read 2/15 of the book on Wednesday

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Two gardens are being fenced in. Both gardens will require the same amount of fencing​ (both gardens have the same perimeter in​
Katyanochek1 [597]

Answer:

The side length of the square is 3 meters.

The side length of the triangle is 4 meters.

Step-by-step explanation:

This is basically a systems of equations word problem.

To solve this, we have to create two different equations. Let's assume s is the side length of the square and t is the side length of the triangle.

We can make the equation 4s=3t, since the perimeters are the same. (A square has 4 sides, a triangle has 3 - multiplying by the side length of each will get perimeter)

We also know that the side length of the triangle is 1 meter longer than the side length of the square, so the equation here becomes t = s+1.

Now, let's substitute the second equation (t = s+1) into the first (4s=3t).

4s = 3(s+1)

Apply the distributive property:

4s = 3s+3

Subtract 3s from both sides:

s=3

So the side length of the square is 3. We can now plug it into the equation t = s + 1 to find the side length of the triangle.

t = 3+1\\\\t=4

Hope this helped!

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3 years ago
Question: %({<520>545]#-[101])
Viktor [21]

Answer:

highest umber is correct answer

4 0
3 years ago
The length of the hypotenuse of a right triangle is 157 units. The length of one leg of the triangle is 132. Lara wrote the foll
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the correct answer for this is the last choice:

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4 0
4 years ago
Read 2 more answers
Find the indicated probability or percentage for the sampling error. The distribution of weekly salaries at a large company is r
Flauer [41]

Answer:

The probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

Step-by-step explanation:

According to the Central Limit Theorem if we have a non-normal population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

\mu=\$1000\\\sigma=\$370\\n=80

As <em>n</em> = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.

Let \bar X represent the sample mean weekly salaries.

The distribution of \bar X is: \bar X\sim N(\$1000,\ \$41.37)

Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.

The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, |\bar X-\mu|= 75.

Compute the probability as follows:

P(-75

                                     =P(-1.81

Thus, the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

3 0
3 years ago
-2x-9y=-25<br> -4X-9y=-23
Arturiano [62]

What are you solving for?

6 0
3 years ago
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