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3 years ago

Math question down below

Mathematics

1 answer:

Kipish [7]3 years ago

3 0

Answer: 6928

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Explanation:

We have two areas we need to find: The area of the trapezoid and the area of the rectangle. Let's call these areas A1 and A2.

Area of Trapezoid = (height)*(base1+base2)/2
A1 = h*(b1+b2)/2
A1 = 80*(150+100)/2
A1 = 80*250/2
A1 = 20000
A1 = 10000

Area of Rectangle = (length)*(width)
A2 = L*W
A2 = 48*64
A2 = 3072

Subtract the two areas (A1-A2) to get the difference D
D = A1 - A2
D = 10000 - 3072
D = 6928

This difference D is exactly equal to the shaded area.

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HELP WITH MATH PLEASE

sergey [27]

Answer:

1) (-3,-2)

2) x=-3

Step-by-step explanation:

1) The vertex of a parabola is the turning point that is the minimum or maximum of the graph, based on the shape of the parabola. In this case, the vertex is the minimum. By looking at the coordinate points, we can tell that the minimum value where the graph changes the direction is at (-3,-2).

2) The line of symmetry is the line on the graph that cuts the shape exactly in half. This means that if you were to fold the graph along this line, the two sides would be identical. All parabolas have a line of symmetry and it always matches the vertex. The equation for the line of symmetry will be x= whatever the x-value of the vertex is. So, for this graph, the line of symmetry is x=-3.

7 0

3 years ago

suppose the nightly rate for a three star hotel in paris is thought to be bell-shaped and symmetrical with a mean of 160 euros a

Westkost [7]

Answer:

95.44%

Step-by-step explanation:

Mean rate (μ) = 160 euros

Standard deviation (σ) = 8 euros

For any given rate, X, the z-score is:

$z=\frac{X-\mu}{\sigma}$

For X = 144 euros:

$z=\frac{144-160}{8}\\ z=-2$

For X= 176 euros

$z=\frac{176-160}{8}\\ z=2$

In a standard distribution, a z-score of -2 corresponds to the 2.28th percentile, while a z-score of 2 corresponds to the 97.72th percentile.

Therefore, the percentage of hotels with rates between 144 and 176 euros is:

$P=97.72-2.28\\P=95.44\%$

8 0

4 years ago

An airplane is preparing to land at an airport. It is 41,300 feet above the ground and is descending at the rate of 3,100 feet p

MissTica

9514 1404 393

Answer:

after 7 minutes
19,600 feet

Step-by-step explanation:

Here's the "pencil and paper" solution:

The two altitude equations are ...

y = 41300 -3100x
y = 2800x

They can be solved by setting the expressions for y equal to each other.

2800x = 41300 -3100x

5900x = 41300

x = 41300/5900 = 7

y = 2800·7 = 19600

The planes will both be at 19,600 feet after 7 minutes.

_____

Attached are solutions from a graphing calculator, and from a calculator app that is able to solve systems of equations.

I find the graphing calculator the easiest to use. I can enter equations using a keyboard, and the solution is displayed in a form that can be copied and pasted.

The calculator app on my phone requires equation entry using a small on-screen keyboard, with multiple key hits required to access some functions. (y is obtained by hitting the x key twice, for example.)

The "pencil and paper" solution is not so difficult, but requires a certain amount of writing (or good short-term memory). The solutions for x and y require separate calculations, whereas the other methods give both x and y at the same time.

3 0

4 years ago

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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

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

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

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

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

3 years ago

Hep blue help please

siniylev [52]

Answer:

31.25%

Step-by-step explanation:

So you do 125 x 100 which gives you 12500, then divide that by 400 to get 31.25.

6 0

3 years ago