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sineoko [7]
3 years ago
15

The back of a park bench is in the shape of a regular trapezoid, as shown in the figure.

Mathematics
2 answers:
ioda3 years ago
7 0

Answer:

a

two triangles with dimensions of 1 ft by 3 ft and a rectangle with dimensions of 4 ft by 3 ft

kap26 [50]3 years ago
3 0

Answer:

the answer is A. two triangles with dimensions of 1 ft by 3 ft and a rectangle with dimensions of 4 ft by 3 ft

Step-by-step explanation:

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plzzz help A model train is on a track 100 cm from its starting point. Turning a switch displaces the train either forward or ba
Ad libitum [116K]

Answer: B.  |x-100| = 25

Explanation:

Draw out a number line. Mark 75, 100 and 125 on the number line as points A, B, and C in that order. Don't worry about the spacing.

The value of x represents where the train is. So if we had say x = 85, then it would be at location 85 on the number line. Writing |x-100| is the distance from x to 100. The absolute value ensures the distance is never negative. We want this distance to be 25 because after traveling 25 cm, the switch is turned, and the train goes the other way.

6 0
4 years ago
Solve using synthetic division.<br><br> (x2 + 3) ÷ (x − 1)
murzikaleks [220]

Answer:

2x+3 / x = 1

Step-by-step explanation:

3 0
3 years ago
Find the area of the shaded region
Mrac [35]

Answer:

40.5

Step-by-step explanation:

Area of square is 9x9=81.

Area of triangle is 1/2(9)(9)=40.5

81-40.5=40.5

40.5 is the area of the shaded region.

4 0
3 years ago
"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt
Westkost [7]

Answer:

a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

d) P(\bar X >28.25)=0.9997=99.97\%

e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

3 0
4 years ago
Estimate the difference between 156 and 31 by first rounding each number to the nearest ten. A.190 B.180 C.120 D.130
Vilka [71]
D is the answer

160 - 30 = 130
4 0
3 years ago
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