The product of 4 and n is -48 translate the sentence into equation

NEED HELP ASAP PLEASE THANK YOU

MariettaO [177]

The answer would be B.

4 0

3 years ago

What are three ways you can solve a proportion?

Komok [63]

Answer:

Vertical.

Horizontal.

Diagonal (often called "cross-products")

Step-by-step explanation:

7 0

3 years ago

What is the midpoint of the segment shown below?

xz_007 [3.2K]

A. (5/2,7/2)

When you look at the graph, giving a rough estimation, 5/2 is 2 1/2 and 7/2 is 3 1/2. The x and y can fit as the midpoint:)

8 0

3 years ago

What is the least common denominator of the rational expressions below? 1/x^2 - 4/3x^2+15x

vladimir1956 [14]

Answer:

The LCD is 3x²

The other two denominators can be factors of 3x²

Step-by-step explanation:

To get the equivalent fractions, divide the original denominator into the LCD, then multiply the original numerator by the quotient.

For the first term: 1/x²

3x² ÷ x² = 3 3×1=3 The equivalent is 3/3x²

Second term: Leave it as it is.

For the third term: 15x

The implied denominator is 1

3x² ÷ 1 = 3x² 3x² × 15 = 135x²

The equivalent is 135x²/3x²

To check the validity, substitute a value for x and calculate the numbers.

For example let x = 3

The original expression becomes

1/3² + 4/3(3²) - 15(3)

1/9 + 4/27 - 45

With the LCD it is

3/27 + 4/27 - 1215/27

5 0

3 years ago

Read 2 more answers

Every day a kindergarten class chooses randomly one of the 50 state flags to hang on the wall, without regard to previous choice

Romashka [77]

Answer:

a) $S=50$

$P(X)=0.02$

b) $P(W,M,C)=8*10^-^6$

c) $P(W_2_3)=1.18*10^-^3$

Step-by-step explanation:

From the question we are told that

Sample space S=50

Sample size n=30

a)Generally the sample space S is

$S=50$

The probability measure is given as

$P(X)=\frac{1}{50}$

$P(X)=0.02$

b)

Generally the probability that the class hangs Wisconsin’s flag on Monday, Michigan’s flag on Tuesday, and California’s flag on Wednesday is mathematically given as

Probability of each one being hanged is

$P(X)=\frac{1}{50}$

Therefore

$P(W,M,C)=\frac{1}{50} *\frac{1}{50}* \frac{1}{50}$

$P(W,M,C)=\frac{1}{125000}$

$P(W,M,C)=8*10^-^6$

c)Generally the probability that Wisconsin’s flag will be hung at least two of the three days is mathematically given as

Probability of two days hung +Probability of three days hung

Therefore

$P(W_2_3)=^3C_2 (1/50) * (1/50) * (49/50) +^3C_3 (1/50) * (1/50) *(1/50)$

$P(W_2_3)=148 / 125000$

$P(W_2_3)=1.18*10^-^3$

3 0

3 years ago