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Vlad [161]
2 years ago
13

Find the area of the following shape.

Mathematics
2 answers:
ad-work [718]2 years ago
5 0

Answer:

57 units^2

Step-by-step explanation:

First find the area of the triangle on the left

ABC

It has a base AC which is  9 units and a height of 3 units

A = 1/2 bh = 1/2 ( 9) *3 = 27/2 = 13.5

Then find the area of the triangle on the right

DE

It has a base AC which is  6 units and a height of 1 units

A = 1/2 bh = 1/2 ( 6) *1  = 3

Then find the area of the triangle on the top

It has a base AC which is  3 units and a height of 3 units

A = 1/2 bh = 1/2 ( 3) *3  = 9/2 = 4.5

Then find the area of the rectangular region

A = lw = 6*6 = 36

Add them together

13.5+3+4.5+36 =57 units^2

nordsb [41]2 years ago
3 0

Answer:

Total Area = 57 sq. units

Step-by-step explanation:

will make it simple and short

Total Area = A1 + A2 + A3

A1 = (7 + 6) * 6/2 = 39 sq. units  (area of a trapezoid)

A2 = 1/2 (9 * 3) = 13.5 sq. units (area of a triangle)

A3 = 1/2 (3 * 3) = 4.5 sq. units  (area of a triangle)

Total Area = 39 + 13.5 + 4.5 = 57 sq. units

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2 years ago
A simple random sample of size n=14 is obtained from a population with μ=64 and σ=19.
madam [21]

Answer:

a) A. The population must be normally distributed

b) P(X < 68.2) = 0.7967

c) P(X  ≥  65.6) = 0.3745

Step-by-step explanation:

a) The population is normally distributed having a mean (\mu_x) = 64  and a standard deviation (\sigma_x) = \frac{19}{\sqrt{14} }

b) P(X < 68.2)

First me need to calculate the z score (z). This is given by the equation:

z=\frac{x-\mu_x}{\sigma_x} but μ=64 and σ=19 and n=14,  \mu_x=\mu=64 and \sigma_x=\frac{ \sigma}{\sqrt{n} }=\frac{19}{\sqrt{14} }

Therefore: z=\frac{68.2-64}{\frac{19}{\sqrt{14} } }=0.83

From z table, P(X < 68.2) = P(z < 0.83) = 0.7967

P(X < 68.2) = 0.7967

c) P(X  ≥  65.6)

First me need to calculate the z score (z). This is given by the equation:

z=\frac{x-\mu_x}{\sigma_x}

Therefore: z=\frac{65.6-64}{\frac{19}{\sqrt{14} } }=0.32

From z table,  P(X  ≥  65.6) =  P(z  ≥  0.32) = 1 -  P(z  <  0.32) = 1 - 0.6255 = 0.3745

P(X  ≥  65.6) = 0.3745

P(X < 68.2) = 0.7967

5 0
3 years ago
Subtract x + 6 from −3x − 4.<br> A) −4x − 10 <br> B) 4x + 10 <br> C) −2x + 2 <br> D) −x + 10
Yuliya22 [10]
The correct answer will be -4x - 10 which is A. X subtracts to the other side and makes it -4x, then 6 +-4 = -10
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3 years ago
A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig
jenyasd209 [6]

Answer:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

n=25 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

Alternative hypothesis:\mu < 20  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

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Zolol [24]

Answer:

12 times (double the times since we are doubling the minutes)

Step-by-step explanation:

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Use cross multiplication

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2 years ago
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