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3 years ago

14

Two students are to be selected at random from a class with 10 girls and 12 boys. what is the probability that both will be girl

s?

1 answer:

Travka [436]3 years ago

7 0

There is an 83% chance that it will be girls

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Can u please find the value of X and both missing angle measures? It’s just notes I was absent when she explained it.

KIM [24]

Heres a tip---- if it has a square its 90 degrees if it doesnt its 180 and if its in an x shape theyre equal

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3 years ago

In this activity, you will help Darin compare the unit prices, in price per pound, of clementines and navel oranges. Clementines

Alisiya [41]

Answer:

The table gives the cost for the number of three-pound bags of clementines. you need to change the number of bags to pounds to be able to get the data for the clementines in units of pounds and cost.

Step-by-step explanation:

each of the bags are three pounds each so 1 bag times 3, 2bags times 3, and then 3bags times 3, hope this answers your question.

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3 years ago

A mechanic works on the overhaul of a transmission. He starts at 8:25 AM and finishes

Mama L [17]

$56.25

8:25 - 4:25 = 4 hours

4 hours - 1 hour for lunch = 3 hours

$18.75 x 3 = $56.25

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2 years ago

Needhdjdjdjdjdjdnhiskdjdjdjdjdjdjjshdjdjdjdjdjdjdjd<br><br>

Lilit [14]

9514 1404 393

Answer:

-(√2)/2

Step-by-step explanation:

The expression evaluated at n=a gives the indeterminate form 0/0, so L'Hopital's rule can be used to find the limit. The second expression comes from differentiating numerator and denominator. Then the form with n=a is no longer indeterminate.

$\displaystyle\lim_{n\to a}{\frac{\sqrt{2n}-\sqrt{3n-a}}{\sqrt{n}-\sqrt{a}}}=\lim_{n\to a}{\frac{\frac{2}{2\sqrt{2n}}-\frac{3}{2\sqrt{3n-a}}}{\frac{1}{2\sqrt{n}}-0}}\\\\=\sqrt{a}\left(\frac{2}{\sqrt{2a}}-\frac{3}{\sqrt{3a-a}}}\right)=\boxed{-\frac{1}{\sqrt{2}}}$

5 0

3 years ago

Sort each equation according to whether it has one solution, infinitely many solutions, or no solution.

Sedaia [141]

Answer:

Equations with one solution: $-2(x-3) = 2x-6$

Equations with no solution:

$6(x+5) = 6x+11\\5(x-2) = 5x-7$

Equations with infinite many solutions: $-3(x-4) = -3x+12$

Step-by-step explanation:

We will solve each equation to see if it has one solution, no solution or infinite many solutions.

An equation where the value of variable can be found has one solution.

An equation with no solution has both sides different.

An equation with infinite solution will have both sides equal after solution(number)

So,

<u>Equation 1:</u>

5(x-2) = 5x-7

$5(x-2) = 5x-7\\5x-10 = 5x-7\\5x-5x = -7+10\\0 = 3$

As both sides are not equal, equation has no solution.

<u>Equation 2:</u>

$-3(x-4) = -3x+12\\-3x+12 = -3x+12\\-3x+3x = -12+12\\0=0\\$

As both sides are equal the equation has infinite many solutions.

<u>Equation 3:</u>

$-2(x-3) = 2x-6\\-2x+6 = 2x-6\\-2x-2x = -6-6\\-4x = -12\\\frac{-4x}{-4} = \frac{-12}{-4}\\x = 3$

As x=3, the equation has only one solution.

<u>Equation 4:</u>

$6(x+5) = 6x+11\\6x+30 = 6x+11\\6x-6x+30 = 11\\30 =11$

The equation has no solution.

Hence,

Equations with one solution: $-2(x-3) = 2x-6$

Equations with no solution:

$6(x+5) = 6x+11\\5(x-2) = 5x-7$

Equations with infinite many solutions: $-3(x-4) = -3x+12$

6 0

2 years ago