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3 years ago

What is m<KNL? Enter your answer in the box

Mathematics

1 answer:

steposvetlana [31]3 years ago

3 0

The measurement of ∠KNL is 102°

First, you must set up an equation to find x.

$5x+2=[3(x+14)$

Then, you must simplify both sides of the equation.

$5x+2=3x+42$

Next, subtract 3x on both sides.

$2x+2=42$

Then, subtract 2 on each side.

$2x=40$

Next, divide 2 on each side.

$x=20$

Now that we have the value of x, we can substitute it in for angle KNL.

$[3(20+14)]$

$[3(34)]$

$102$

$102$ °

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When does it reach maximum height? H(t)=-16t^2+112t+288

ivolga24 [154]

Answer:

484 feet maximum

Step-by-step explanation:

H(t)=-16t^2+112t+288

The maximum height occurs at the vertex of this graph.

Let us find the vertex.

vertex x-coordinate = -b/2a

x = -b/2a = -(112)/ (2*-16) = 7/2 sec.

H(7/2) = - 16 *(7/2)^2 + 112*(7/2) + 288

H(7/2) = -196 + 392 + 288 = 484 feet ..

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3 years ago

three different accounts are described below. order the accounts to their values after 10 years, from greatest to least.

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Step-by-step explanation:

From the close eye, this seems to me the best and most factual answer to be found!! Goodluck with your quiz!

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3 years ago

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The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.

Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5 {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

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

where, $\bar X$ = sample mean MCAT score = 32.2

s = sample standard deviation = 4.2

n = sample of students = 40

So, the test statistics = $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$ ~ $t_3_9$

= 4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

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3 years ago

If x does not equal 16 then 2x+10 does not equal 42

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A) 19

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What is m<KNL? Enter your answer in the box​

What is m<KNL? Enter your answer in the box