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3 years ago

9.

Mathematics

1 answer:

alexgriva [62]3 years ago

6 0

That is the answer because the only thing that changes is the multiplication but either way it will multiply to the same answer!

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Giving Brainliest, hard question ,pretty sure nobody knows this

topjm [15]

south side of town? I think

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3 years ago

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HELP AGAIN PLEASE I TRIED IT ON MY OWN AND FAILED :(

Alex787 [66]

Answer: ∠ J = 62° , ∠ K = 59° , ∠ L = 59°

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

It is given that it is an Isosceles Triangle, where L J ≅ K J

It follows that ∠ K ≅ ∠ L

⇒ 5x + 24 = 4x + 31

⇒ x + 24 = 31

⇒ x = 7

Input the x-value into either equation to solve for ∠ K & ∠ L:

∠ K = 5x + 24

= 5(7) + 24

= 35 + 24

= 59

∠ K ≅ ∠ L ⇒ ∠ L = 59

Next, find the value of ∠ J:

∠ J + ∠ K + ∠ L = 180 Triangle Sum Theorem

∠ J + 59 + 59 = 180

∠ J + 118 = 180

∠ J = 62

8 0

3 years ago

What is the ratio of rise to run between the points (0, 4) and (5, 6)? A.5/2 B.-5/2 C.2/5 D.-2/5

RUDIKE [14]

Answer:

http://mansfieldalgebra1.weebly.com/uploads/1/0/4/4/10447339/rc_2_answer_key.pdf

Step-by-step explanation:

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3 years ago

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Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.6, \sigma = 0.5$ .

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when $X = 10$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 9.6}{0.5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8 - 9.6}{0.5}$

$Z = -3.2$

$Z = -3.2$ has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have $n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1$ .

This probability is 1 subtracted by the pvalue of Z when $X = 9.8$ . So:

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.8 - 9.6}{0.1}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

5 0

3 years ago

What basic trigonometric identity would you use to verify that

Dafna1 [17]

Given the equation:

$\frac{\sin^2x+\text{cos}^2x}{\cos x}=\sec x$

Let's determine the trigonometric identity that you could be used to verify the exquation.

Let's determine the identity:

Apply the trigonometric identity:

$\sin ^2x+\cos ^2x=1$

$\cos x=\frac{1}{\sec x}$

Replace cosx for 1/secx

Thus, we have:

$\begin{gathered} \frac{\sin^2x+\cos^2x}{\frac{1}{\sec x}} \\ \\ =(\sin ^2x+\cos ^2x)(\sec x) \\ \text{Where:} \\ (\sin ^2x+\cos ^2x)=1 \\ \\ We\text{ have:} \\ (\sin ^2x+\cos ^2x)(\sec x)=1\sec x=\sec x \end{gathered}$

The equation is an identity.

Therefore, the trignonometric identity you would use to verify the equation is:

$\cos ^2x+\sin ^2x=1$

ANSWER:

$\cos ^2x+\sin ^2x=1$

6 0

2 years ago