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xenn [34]
3 years ago
10

3( c - 4 )= I need help some one please help me

Mathematics
1 answer:
AysviL [449]3 years ago
3 0

Answer:

3c - 12

Step-by-step explanation:

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The perimeter of a playing field for a certain sport is 214 ft. The field is a rectangle, and the length is 41 ft longer than th
malfutka [58]

Step-by-step explanation:

Perimeter of a rectangle = 2(L+W)

L is the length of the rectangle:

W is the width of the rectangle.

Given

Perimeter = 214feet

If the length is 41 ft longer than the width, then L = 41+W

Substitute L = 41+W into the formula:

P = 2(L+W)

214 = 2(41+W+W)

214 = 2(41+2W)

214 = 82+4W

4W = 214-82

4W = 132

W = 132/4

W = 33 feet

Since L = 41+W

L = 41+33

L = 74 feet

Hence the dimension of the field is 74ft by 33ft

<em>The width of the playing field is 33feet</em>

8 0
3 years ago
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The answer is option C. 9xy sqrt 2x
5 0
3 years ago
PLEASE HELP ME! I will mark you as a BRAINLIST!!
disa [49]
The answer is  -15

See the attached image for the steps on how I got that answer.

8 0
3 years ago
Simplify each expression <br>(mx+2ny+z)<br> and<br>(y^3-zx^3)(-yz)<br>AND<br>4xy^3(x^4y-5x)
Sliva [168]

Answer:  mx+2ny+z               −zy4−zx3           4xy3x4y−5x

 


Step-by-step explanation:

there u go

8 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 149 millimeters,
Artemon [7]

Answer:

The probability that the sample mean would differ from the population mean by more than 0.5 millimeters is 0.2420.

Step-by-step explanation:

Let <em>X</em> = the diameter of the steel bolts manufactured by the steel bolts manufacturing company Thompson and Thompson.

The mean diameter of the bolts is:

<em>μ</em> = 149 mm.

The standard deviation of the diameter of bolts is:

<em>σ</em> = 5 mm.

A random sample, of size <em>n</em> = 49, of steel bolts are selected.

The population of the diameter of bolts is not known.

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

The mean of the sampling distribution of sample mean is:

\mu_{\bar x}=\mu=149\ mm

The standard deviation of the sampling distribution of sample mean is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{49}}=0.7143

Compute the probability that the sample mean would differ from the population mean by more than 0.5 millimeters as follows:

P(\bar X>\bar x)=P(\frac{\bar X-\mu_{\bar x}}{\sigma _{\bar x}}>\frac{0.50}{0.7143})\\=P(Z>0.70)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 0.5 millimeters is 0.2420.

8 0
4 years ago
Read 2 more answers
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