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2 years ago

What is the value of w? -8w + 5= 45

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

7 0

Subtract 5 from 45 which equals 40 so -8w can be alone. Then divide 40 by -8w you will get -5.

w=-5

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What are the solutions to the equation?

Maru [420]

rearrange the equation to where you set it equal to 0 by moving the 25 over.

n2 - 8n -9 = 0

now factor

(n-9)(n+1)

so n1 = 9

n2 = -1

6 0

3 years ago

Kendra just bought a new house and needs to buy new sod for her backyard. If the dimensions of her yard are 24.6 feet by 14.8 fe

Serhud [2]

Given:

Dimensions of Kendra's yard are 24.6 feet by 14.8 feet.

To find:

The area of her yard.

Solution:

Let as consider,

Length of yard = 24.6 ft

Width of yard = 14.8 ft

We know that,

Area of a rectangle = Length × Width

Using the above formula, we get the area of her yard.

$Area=24.6\times 14.8$

$Area=364.08$

Area of her yard is 364.08 ft² and without including the units ft², we get 364.08.

Therefore, the area of her yard is 364.08.

6 0

3 years ago

The output of a process is stable and normally distributed. If the process mean equals 23.5, the percentage of output expected t

erastovalidia [21]

Answer:

Option a) 50% of output expected to be less than or equal to the mean.

Step-by-step explanation:

We are given the following in the question:

The output of a process is stable and normally distributed.

Mean = 23.5

We have to find the percentage of output expected to be less than or equal to the mean.

Mean of a normal distribution.

The mean of normal distribution divides the data into exactly two equal parts.

50% of data lies to the right of the mean.

50% of data lies to the right of the mean

Thus, by property of normal distribution 50% of output expected to be less than or equal to the mean.

6 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0

3 years ago

Find each percent increase round to the nearest percent from $5 to $8

asambeis [7]

First subtract the two numbers:

8 - 5 = 3

Now divide this to the original:

3 / 5 = 0.6

Multiply by 100:

0.6 * 100 = 60%

7 0

3 years ago