Find the perimeter of this shape.

. Show ALL work for the following problem... if you write the answer only, you will not get brainlist: 2(x - 2) = 4x - 6(x - 2)

Rom4ik [11]

Distribute 2 to x and -2, and -6 to x and -2

2(x - 2) = 2x - 4

-6(x - 2) = -6x + 12

2x - 4 = 4x - 6x + 12

Simplify. Combine like terms

2x - 4 = (4x - 6x) + 12

2x - 4 = -2x + 12

Isolate the x. Add 4 to both sides, and 2x to both sides

2x (+2x) - 4 (+4) = -2x (+2x) + 12 (+4)

2x + 2x = 12 + 4

Simplify

4x = 16

Isolate the x. Divide 4 from both sides

4x/4 = 16/4

x = 16/4

x = 4

4 is your answer for x.

hope this helps

6 0

3 years ago

What is 1x+3y=18 in slope intercept form

photoshop1234 [79]

Answer:

y = -x/3 + 6

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Standard Form: Ax + By = C

Slope-Intercept Form: y = mx + b

m - slope
b - y-intercept

Step-by-step explanation:

Step 1: Define

Standard Form: x + 3y = 18

Step 2: Rewrite

Find slope-intercept form

Subtract x on both sides: 3y = -x + 18
Divide 3 on both sides: y = -x/3 + 6

3 0

3 years ago

Read 2 more answers

Suppose that scores on an aptitude test are normally distributed with a mean of 100 and a standard deviation of 4.6. Scores on a

DerKrebs [107]

Answer:

The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.

Step-by-step explanation:

1. Let's check all the information given to us to answer the question correctly:

Mean of the scores on the aptitude test = 100

Standard deviation of the aptitude test = 4.6

Felix's score on the aptitude test = 106

Mean of the scores on the knowledge test = 70

Standard deviation of the knowledge test = 2.4

Felix's score on the knowledge test = 75

2. Which statement best describes Felix's scores on the two tests comparatively?

Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.

Therefore, a score of 106 on the aptitude test, will have the following z-score:

106 - 100 = 6 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

6/4.6 = 1.30

A score of 75 on the knowledge test, will have the following z-score:

75 - 70 = 5 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

5/2.4 = 2.08

The z-scores of Felix were 1.30 on the aptitude test and 2.08 on the knowledge test. He performed better on the aptitude test because 2.08 > 1.30, so the correct statement that best describes Felix's scores on the two tests comparatively is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.

6 0

3 years ago

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

3 years ago

The security system in a house has two units that set off an alarm when motion is detected. Neither one is entirely reliable, bu

Mariulka [41]

Answer:

Probability that detector B goes off is '0.615'

Step-by-step explanation:

Given that:

1) Probability that detector A goes off and detector B does not go off is 0.25.

2)Probability that detector A does not go off is 0.35.

3)Probability that detector A goes off is (1-0.35)=0.65

Assuming that

Probability that detector B goes off is 'p' Hence the probability that detector B does not goes off is (1-p)

Thus the probability that detector A goes off and detector B does not go off is product of the individual probabilities

$P(E)=0.65\times (1-p)\\\\0.65\times (1-p)=0.25\\\\0.65-0.65p=0.25\\\\0.65-0.25=0.65p\\\\\therefore p=\frac{0.40}{0.65}=0.615$

Probability that detector B goes off is '0.615'

5 0

3 years ago