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4 years ago

Math help please....

Mathematics

1 answer:

Alex_Xolod [135]4 years ago

7 0

Let's solve your equation step-by-step.−7(x−4)+8x=12+14

Step 1: Simplify both sides of the equation.−7(x−4)+8x=12+14

Simplify: (Show steps)x+28=26

Step 2: Subtract 28 from both sides.x+28−28=26−28

x=−2

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One side of trapezoid is 10 inches. The side parallel to this is twice as long as this side. The height is half as long as the g

Vika [28.1K]

The formula for area of a trapezoid is: 1/2 (b1+b2) h
b1=10
b2=10·2=20 because it stated the side parallel is twice as long as the side above
h=10/2=5 because the height is half as long as the give side length
area of the trapezoid is: 1/2 (10+20) 5
(5·10)·5
50·5
area of the trapezoid is 250 in²

4 0

3 years ago

JUST ANSWER PLEASE!!! QUICK

jeka57 [31]

Answer:

Options 1 and 3

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given: the solution for the following system is (2,9)

Px + Qy = R ⇒(1)

Tx + Uy = V ⇒(2)

We will check which system of equation has the same solution.

System A) Px + Qy = R

(P+T)x + (Q+U)y = R+V ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

System B) Px + Qy = R

(P+2T)x + (Q+2U)y = R-2V ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

(P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

System C) (T-P)x + (U-Q)y = V-R ⇒(5)

Tx + Uy = V

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

System D) (T-P)x + (Q+U)y = V-R ⇒(6)

Tx + Uy = V

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) OR (T-P) and (Q-U)

So, system D has not the same solution of (2,9)

System E) (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

Tx + Uy = V

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

Options 1 and 3

5 0

3 years ago

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal prob

Novay_Z [31]

Answer:

a) By the Central Limit Theorem, it is approximately normal.

b) The standard error of the distribution of the sample mean is 1.8333.

c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours

e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

Step-by-step explanation:

To solve this question, 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 estabilishes 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 $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

Mean of 36 hours and a standard deviation of 5.5 hours.

This means that $\mu = 36, \sigma = 5.5$

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

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that $n = 9$ . So

$s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333$

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

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

By the Central Limit Theorem

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

$Z = \frac{38 - 36}{1.8333}$

$Z = 1.09$

$Z = 1.09$ has a pvalue of 0.8621

1 - 0.8621 = 0.1379

0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

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

$Z = \frac{34.5 - 36}{1.8333}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061.

1 - 0.2061 = 0.7939

0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.

e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?

pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

4 0

3 years ago

What does X equal?? Please help me out!!

Alja [10]

X = 80 + 20 = 100 degrees

3 0

3 years ago

National statistics show that 23% of men smoke and 18.5% of women smoke. A random sample of 175 men indicated that 45 were smoke

NeTakaya

Answer:

98% C.I. = [-0.1270, 0.0994]

Step-by-step explanation:

Here

p1= proportion of men smokers

p1= 45/175= 0.2571

p2= proportion of women smokers

p2= 42/155= 0.2709

1) Choose the significance level ∝= 0.02

2) The formula for confidence interval is

(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

and

(p1^- p2^) z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

3) The z value is 2.05

4) Calculations:

(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

Putting the values

= 0.2571-0.2709)(-2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)

=-0.1270 ( using calculator)

And

(p1^- p2^)z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

Putting the values

= 0.2571-0.2709)(2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)

=0.0994 ( using calculator)

and Putting ± 2.05 in the confidence interval formula we get limits (-0.1270, 0.0994)

6 0

3 years ago