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iren [92.7K]
3 years ago
9

I have a questions does any take the 3 chapter test about math if any one please say yes in the comments below

Mathematics
1 answer:
Anastaziya [24]3 years ago
3 0
What? Not everybody has the same math book, try asking someone in your school
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The radius of a circle is half the length of the circumference. 
So multiply 12 by itself and you get 24.
The answer is C.
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If a group of independent variables are not significant individually but are significant as a group at a specified level of sign
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Answer:

D) collinearity

Step-by-step explanation:

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3 years ago
M+1 3/8=5 What does m equal?
Arte-miy333 [17]

\text{ The value of m is equal to 3.625 or }\frac{29}{8} \text{ or } 3\frac{5}{8}

<em><u>Solution:</u></em>

Given that we have to find the value of "m"

Given expression is:

m + 1\frac{3}{8} = 5

<em><u>Let us first convert the mixed fraction to improper fraction</u></em>

Steps to follow:

Divide the numerator by the denominator.

Write down the whole number answer.

Then write down any remainder above the denominator.

Therefore,

1\frac{3}{8}=\frac{8 \times 1+3}{8} = \frac{8+3}{8} = \frac{11}{8}

Now the expression becomes,

m + \frac{11}{8} = 5

Keep the variable "m" on L.H.S and move the constant to R.H.S

m = 5 - \frac{11}{8}\\\\m = \frac{5 \times 8 -11}{8}\\\\m = \frac{40-11}{8}\\\\m = \frac{29}{8}\\\\\text{In mixed fractions, we can write as }\\\\m = \frac{29}{8} = 3\frac{5}{8}

Thus value of m is equal to 3.625 or \frac{29}{8} \text{ or } 3\frac{5}{8}

3 0
3 years ago
A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
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