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3 years ago

What is the midpoint

Mathematics

1 answer:

liq [111]3 years ago

5 0

Answer:

midpoint formula:

x1+x2/2 y1+y2/2

-4 + (-2)/2 which then simplifies to -6/2 and then to -3

2 + 2 /2 which then simplifies to 2

x= -3

y= 2

so (-3,2) is the coordinate point of the midpoint of this segment

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Cos 35º sin 55° + cos 55° sin 35°

Tresset [83]

Answer:

Step-by-step explanation:

CosASinB + SinACosB

Sin(A+B)

Sin(35+55) = Sin(90) = 1

5 0

3 years ago

Read 2 more answers

Simplify for a lot of points<br> (1-sin^2x)(1-tan^2x)

notka56 [123]

1/sin^2x-1/tan^2x=
1/sin^2x-1/ (sin^2x/cos^2x)<<sin tan= sin/cos>>
= 1/sin^2x- cos^2x / sin^2x
= (1- cos^2x) / sin^2x <<combining into a single fraction>>
sin^2 x / sin^2x <<since 1- cos^2 x sin^2 x

=1
this simplifies to 1.

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4 years ago

Slope 4.2; y-intercept (0,3.4)

umka2103 [35]

Answer:I don't know if you're asking for equation or not cause you didn't mention the question so I'm just gonna write down the equation.

y=4.2x+3.4

Use this formula➡️y=mx+b

m is slope, b is y-intercept , x and y are the points like (0, 3.4)

You said it's the y-intercept but it's not. it's just a point on the graph

Step-by-step explanation:

5 0

4 years ago

Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut

defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .