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zhenek [66]
3 years ago
6

Solve the equation. –3x + 1 + 10x = x + 4

Mathematics
1 answer:
Sedaia [141]3 years ago
6 0
<span>–3x + 1 + 10x = x + 4
Subtract x from both sides
-4x + 1 + 10x = 4
Add 10x to -4x
6x + 1 = 4
Subtract 1 from both sides
6x = 3
Divide both sides by 6
Final Answer: x= 0.5</span>
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3 years ago
23=x-9;x+14 I need help with this problem.
frez [133]
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6 0
2 years ago
You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

8 0
3 years ago
Read 2 more answers
Two altitudes of a triangle have lengths $12$ and $14$. What is the longest possible integer length of the third altitude
nataly862011 [7]

The longest possible altitude of the third altitude (if it is a positive integer) is 83.

According to statement

Let h is the length of third altitude

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.

From Area of triangle

A = 1/2*B*H

Substitute the values in it

A = 1/2*a*12

a = 2A / 12 -(1)

Then

A = 1/2*b*14

b = 2A / 14 -(2)

Then

A = 1/2*c*h

c = 2A / h -(3)

Now, we will use the triangle inequalities:

  • a < b+c

2A/12 < 2A/14 + 2A/h

Solve it and get

h<84

  • b < a+c

2A/14 < 2A/12 + 2A/h

Solve it and get

h > -84

  • c < a+b

2A/h < 2A/12 + 2A/14

Solve it and get

h > 6.46

From all the three inequalities we get:

6.46<h<84

So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.

Learn more about TRIANGLE here brainly.com/question/2217700

#SPJ4

8 0
2 years ago
Please help please...
Leya [2.2K]
First one is
(2x+5)

Second one is
“All of these are correct”

Hope this helped
8 0
3 years ago
Read 2 more answers
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