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Lelu [443]
3 years ago
12

If AB=CB then B must be the midpoint of AC

Mathematics
1 answer:
liq [111]3 years ago
5 0

This answer is true, any other questions may I help you


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A map has a scale of 2.5 inches=125 miles. If the actual distance is 300 miles what is the distance on the map
e-lub [12.9K]
Answer 5.5

Explanation 125 + 125= 250
250 + 50=300

Is 2 125s and 1 50

Each 125 is 2.5

But in this case there is 2 125s

2.5+2.5=5
plus the other half = 5.5

BRAINLIST?
4 0
2 years ago
Help plzzzzzzzzzzzzz
romanna [79]

Answer:

C

Step-by-step explanation:

i just did sum hw work on this like a hr ago

6 0
3 years ago
An educator wants to determine whether a new curriculum significantly improves standardized test scores for fourth grade student
vesna_86 [32]

Answer:

Independent Sampling

Step-by-step explanation:

There are two scenarios for independent sampling .

Testing the mean we get the differences between samples from each population. When both samples are randomly  inferences about the populations. can get.

Independent sampling are sample that are selected randomly. Observation does not depend upon value.Many analysis assume that sample are independent.

In this statement 90 dash are divides into two groups Group 1 and Group 2 . Both are standardized that mean both are randomly selected. Means are observed. Observation doesn't depend upon value.  So this style of sampling is independent Sample.

5 0
3 years ago
Solve the equation.<br> (10x + 5) + (6x − 1) = 180<br><br> A. 9<br> B. 11<br> C. 17<br> D. 23
cluponka [151]
(10x+5) + (6x-1)=180
10x+5+6x-1=180
16x+4=180
16x+4=180
16x=180-4
16x=178
x=11
3 0
3 years ago
Read 2 more answers
Astronomers treat the number of stars X in a given volume of space as Poisson random variable. The density of stars in the Milky
cupoosta [38]

Answer:

Explained below.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of stars in a given volume of space.

X\sim \text{Poisson}\ (\lambda=1)

The probability mass function of <em>X</em> is:

p_{X}(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}

(7)

Compute the probability of exactly two stars in 16 cubic light-years as follows:

P(X=2)=\frac{e^{-1}\times 1^{2}}{2!}=\frac{e^{-1}}{2}=\frac{0.36788}{2}=0.18394\approx 0.184

(8)

Compute the probability of three or more stars in 16 cubic light-years as follows:

P(X\geq 3)=1-P(X

(9)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in 4 cubic light years there will be, 4 × (1/16) = (1/4) stars.

(10)

In 16 cubic light years there is only 1 star.

Then in 1 cubic light years there will be, (1/16) stars.

Then in <em>t</em> cubic light years there will be, [<em>t</em> × (1/16)] stars.

7 0
3 years ago
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