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2 years ago

Please Help Me Answer These

Mathematics

1 answer:

charle [14.2K]2 years ago

8 0

Answer:

Question 1 : -2 3/4 × 3 1/5 . Change the fractions to improper fraction and multiply .

= -11/4 × 16/5

= -176/20

= -44/5

= -8 4/5

<h3> Question 1 answer : -8 4/5</h3>

Question 2: -4.6/-2

<h3>Question 2 answer : 2.3</h3>

Question 3: - 7/8 ÷ (-2)(-4)

<h3>Question 3 answer : -7/4</h3>

Question 4 : -4 27/30

<h3>Question 4 answer : c</h3>

Question 5 : -3 13/20

<h3>Question 5 answer : -3.65</h3>

<h3>Therefore The answer are </h3><h3>-8 1/4 , 2.3 , -7/4 , c , -3.65</h3>

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What’s is the difference between an improper fraction and mixed number

Darya [45]

Answer:

An improper fraction is always 1 or greater than 1. And, finally, a mixed number is a combination of a wholenumber and a proper fraction.In a proper fraction, the numerator is always less than the denominator. Examples ofproper fractions include and

6 0

2 years ago

May someone please help me with this :)

AlladinOne [14]

Answer: 16 pi or ≈50.27

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2 years ago

Help with this please

Svet_ta [14]

The answer to the question is D

4 0

3 years ago

Read 2 more answers

In triangle ABC, m<A = 30° and m

Nataliya [291]

Answer:

100°

Step-by-step explanation:

The sum of the angles in a triangle is 180°

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8 0

2 years ago

Consider a game in which a participant pays $2 to roll a die. The participant receives $3 if they roll a 1 (i.E. They go up by a

Sauron [17]

Answer:

The expected monetary value of a single roll is $1.17.

Step-by-step explanation:

The sample space of rolling a die is:

S = {1, 2, 3, 4, 5 and 6}

The probability of rolling any of the six numbers is same, i.e.

P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = $\frac{1}{6}$

The expected pay for rolling the numbers are as follows:

E (X = 1) = $3

E (X = 2) = $0

E (X = 3) = $0

E (X = 4) = $0

E (X = 5) = $0

E (X = 6) = $4

The expected value of an experiment is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected monetary value of a single roll as follows:

$E(X)=\sum x\cdot P(X=x)\\=[E(X=1)\times \frac{1}{6}]+[E(X=2)\times \frac{1}{6}]+[E(X=3)\times \frac{1}{6}]\\+[E(X=4)\times \frac{1}{6}]+[E(X=5)\times \frac{1}{6}]+[E(X=6)\times \frac{1}{6}]\\=[3\times \frac{1}{6}]+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]\\+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]+[4\times \frac{1}{6}]\\=1.17$

Thus, the expected monetary value of a single roll is $1.17.

7 0

3 years ago