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ehidna [41]
3 years ago
15

Given that 5 x : 8 = 7 : 2 Calculate the value of x .

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

x = 5.6

Step-by-step explanation:

5x:8 = 7:2

First, find the relationship between 2 and 7.

2 x 3.5 = 7

Since the ratios are equal, this means:

8 x 3.5 = 5x

Solve algebraically. First simply 8 x 3.5

5x = 28

Divide both sides by 5

x = 5.6

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A region is bounded by y=e−3x, the x-axis, the y-axis and the line x = 3. If the region is the base of a solid such that each cr
grigory [225]

Answer:

0.333\pi units³

Step-by-step explanation:

Think process:

The equation is given as y = e^{-3x}

Let, y = f (x)

Therefore, f (x) = e^{-3x}

We know that the limits are y-axis and x= 3

Y-axis: x= 0

then limits are given as x= 0 and x = 3

Integrating  gives:

\int\limits^3_0 {e^{-3x} } \, dx = \frac{-1}{3}e^{-3x} + C

calculating from x= 0 to x = 3, we know volume is given by \pi \int\limits^a_b {f(x)} \, dx

= \pi[ \frac{-1}{3} e^{-9}  - (\frac{-1}{3} e^{0})]

= \pi[0.000041136 + 1/3]

= 0.333\pi units³

8 0
3 years ago
Slope =-1, y-intercept = 3
ss7ja [257]

If you are looking for the equation, then it would be y = -x+3

This is the same as y = -1x+3 which is in slope intercept form y = mx+b

m = -1 = slope

b = 3 = y intercept

If you want to graph this, then plot the points (0,3) and (1,2). Then draw a straight line through them both. Extend the line in both directions as much as possible.

7 0
2 years ago
Read 2 more answers
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
2 years ago
Will mark brainiest! !!50 Points!! The first question since it got cut off is : (A) is the sequence arithmetic, geometric, or ne
blagie [28]

Answer:

Geometric

Step-by-step explanation:

8 0
2 years ago
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3.
nasty-shy [4]

Answer:

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Step-by-step explanation:

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