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2 years ago

Please help me with this question

Mathematics

1 answer:

Bad White [126]2 years ago

8 0

2/7 = 10/35
3/5 = 15/35
Total = 25/35
Total = 5/7

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What does this mean? evaluate the expression for y=54y+10

salantis [7]

We have the expression:

$4y+10$

And want to evaluate in y=5, this means we have to replace y in the expression for 5 and calculate the value of the expression, so:

$\begin{gathered} 4y+10,y=5 \\ 4\cdot5+10 \\ 20+10 \\ 30 \end{gathered}$

The expression is 30 when y=5.

7 0

10 months ago

Please help 6TH GRADE MATH HELP ME OUT PLEASE

wel

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▹ Answer

▹ Step-by-Step Explanation

$\frac{1}{4} + \frac{1}{10} \\= Common denominator is 20\\\\\frac{1}{4} = \frac{5}{20} \\\\\frac{1}{10} = \frac{2}{20} \\\\\frac{5}{20} + \frac{2}{20} \\\\= \frac{7}{20}$

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7 0

3 years ago

Read 2 more answers

Please help and show ur work:)

Sidana [21]

3/4...there’s no work to show

6 0

3 years ago

What are the coordinates of the midpoint for segment AB if A is (14 , -2) and B is (7 , -8)

Alenkasestr [34]

Answer:

Step-by-step explanation:

(14 + 7)/2 = 21/2= 10.5

(-2 + (-8))/2= -10/2= -5

Midpoint(10.5, -5)

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2 years ago

Dustin has a bag of marbles with 4 blue marbles, 4 white marbles, and 3 red marbles.

diamong [38]

Answer:

$\dfrac{6}{55}$

$\dfrac{4}{165}$

Step-by-step explanation:

Dustin has a bag of marbles with 4 blue marbles, 4 white marbles, and 3 red marbles. There are 11 marbles in total.

A. The probability that the first marble drawn is blue marble is

$\dfrac{4}{11},$

the probability that the second marble drawn is red marble is

$\dfrac{3}{10}$

The probability that the first marble drawn is blue and the second is red is

$\dfrac{4}{11}\cdot \dfrac{3}{10}=\dfrac{12}{110}=\dfrac{6}{55}$

B. The probability that the first marble drawn is red marble is

$\dfrac{3}{11},$

the probability that the second marble drawn is white marble is

$\dfrac{4}{10}$

The probability that the first marble drawn is red and the second is white is

$\dfrac{3}{11}\cdot \dfrac{4}{10}=\dfrac{12}{110}=\dfrac{6}{55}$

C. The probability that the first marble drawn is blue marble is

$\dfrac{4}{11},$

the probability that the second marble drawn is blue marble is

$\dfrac{3}{10},$

the probability that the third marble drawn is blue marble is

$\dfrac{2}{9},$

The probability that the first, second and third marbles drawn are blue is

$\dfrac{4}{11}\cdot \dfrac{3}{10}\cdot \dfrac{2}{9}=\dfrac{24}{990}=\dfrac{4}{165}$

8 0

2 years ago