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alukav5142 [94]
3 years ago
10

Second part of the problem find the x. Link: https://brainly.com/question/26290433

Mathematics
1 answer:
Sliva [168]3 years ago
3 0

Answer:

no

Step-by-step explanation:

it does not work r÷eeeeeeeeeeeeeeee eeeeeeeeeeeeeeee eeeeeeeeeeeeeeee

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Help Asap Immediately!!
Svetlanka [38]

Answer:

its 432

Step-by-step explanation:

You multiply the base times the side length then divide by 2. hope this helps

3 0
3 years ago
(PLEASE HELP) Triangle LMN is similar to triangle OPR. Find the measure of side RO. Round your answer to the nearest tenth if ne
AysviL [449]

Answer:

<h3>RO = 107.36</h3>

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

ΔLMN = ΔOPR ----- (<em>given</em>)

<h3>so,-----------------------</h3>

NM / NL = RP / RO

19 / 34 = 60 / RO

RO = 60 × 34 ÷ 19

RO = 107.36

6 0
3 years ago
HELP ASAP I'M FAILING AND I DON'T UNDERSTAND THIS
nirvana33 [79]
The intercept can be found when all other variables are equated to zero.
x-intercept when y = 0 and z = 0: 8x + 6*0 + 3*0 = 24 gives x = 3
y-intercept when x = 0 and z = 0: 8*0 + 6y + 3*0 = 24 gives y = 4
z-intercept when x = 0 and y = 0: 8*0 + 6*0 + 3z = 24 gives z = 8

The intercepts are (3, 0, 0), (0, 4, 0), and (0, 0, 8).
7 0
3 years ago
a machine takes 4.2 hours to make 7 parts. at that rate how many parts can the machine make in 28.8 hours
Karo-lina-s [1.5K]
In 1 hour, a machine could make:
7 \div 4.2
parts
In 28.8 hours, that machine make
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6 0
3 years ago
Read 2 more answers
Suppose that in a random sample of 300 employed Americans, there are 57 individuals who say that they would fire their boss if t
otez555 [7]

Answer:

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 300, \pi = \frac{57}{300} = 0.19

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 - 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.1456

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 + 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.2344

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

4 0
3 years ago
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