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Free_Kalibri [48]
2 years ago
8

Find 985 + 659 this is freclke btw

Mathematics
2 answers:
ozzi2 years ago
6 0

Answer:

if your just adding then it would be 1644.

Step-by-step explanation:

Or are you doing something different?

anygoal [31]2 years ago
5 0

Answer:1700

Step-by-step explanation:magic

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Answer:

help me pls

Step-by-step explanation:

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3 years ago
Help me solve this two questions
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Answer:

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3 years ago
If the first step in the solution of the equation 2x - 8 = 5x + 3 is "subtract 2x," then in the form of a paragraph, explain in
monitta

x = \frac{-11}{3}

<u>Step-by-step explanation:</u>

Here we have , If the first step in the solution of the equation 2x - 8 = 5x + 3 is "subtract 2x," then in the form of a paragraph, explain in complete sentences the next steps necessary to completely solve the equation for x. Let's solve this :

2x - 8 = 5x + 3

⇒ 2x - 8 = 5x + 3

⇒ (2x - 8)-2x = (5x + 3)-2x { subtract 2x from both side }

⇒ 0 - 8 = 5x-2x + 3

⇒ - 8 = 3x + 3

⇒ - 8 -3= 3x + 3-3 { subtract 3 from both sides }

⇒ - 11 = 3x + 0

⇒ - 11 = 3x

⇒ \frac{-11}{3}  = \frac{3}{3} x { divide 3 from both sides }

⇒ x = \frac{-11}{3}

3 0
3 years ago
Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas
polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 120-1 = 119

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have T = 1.6578.

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

s = \frac{650}{\sqrt{120}} = 59.34

Now, we multiply T and s

M = T*s = 59.34*1.6578 = 98.37

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that \alpha = 0.95

So

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

With 119 and 0.025 in the t-distribution table, we have T = 1.9801.

M = T*s = 59.34*1.9801 = 117.50

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that \alpha = 0.99

So

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005

With 119 and 0.025 in the t-distribution table, we have T = 2.6178.

M = T*s = 59.34*2.6178 = 155.34

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0
3 years ago
What is the midpoint of the segment shown below
Softa [21]

Answer:

(-1, -1/2)

Step-by-step explanation:

To find the x coordinate of the midpoint, add the x coordinates and divide by 2

( -11+9)/2 = -2/2 = -1

To find the y coordinate of the midpoint, add the y coordinates and divide by 2

(0+-1)/2 = -1/2

(-1, -1/2)

6 0
3 years ago
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