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3 years ago

11

Rosa bought d drawings to and art show. After selling 14 of them she had 34 left. Identify the equation that represents this sit

uation and the correct solution

1 answer:

Alexxx [7]3 years ago

7 0

Answer:

14+34=d hope this helps

You might be interested in

A. Use substitution and show work

Alja [10]

A) 2x -3(3x -7) = 7 . . . . substitute for y
.. -7x = -14 . . . . . . . . . . . subtract 21
.. x = 2 . . . . . . . . . . . . . divide by -7
.. y = 3*2 -7 = -1 . . . . . . substitute for x
(x, y) = (2, -1)

B) 5(5x -2y) -2(3x -5y) = 5(8) -2(1) . . . . subtract 2 times 2nd equation from 5 times 1st
.. 19x = 38 . . . . simplify
.. x = 2 . . . . . . . divide by 19
.. y = (5x -8)/2 = 1 . . . . solve 1st equation for y, substitute x
(x, y) = (2, 1)

C) See the first graph.
(x, y) = (-4, 1)

D) See the second graph.
(x, y) = (3, 2)

6 0

3 years ago

The length of the rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 f

Temka [501]

Let's first say that L=W+44

and then remember that perimeter is P=2L+2W

replace the L with W+44

we then get P=2(W+44)+2W, now I'll solve it

P=2W+88+2W
P=4W+88

substitute 288 for P

288=4W+88
200=4W
50=W

so now we now how wide the court is. add 44 to find the length which gives you L=94

as always plug the numbers back into your perimeter equation to ensure L and W are correct

8 0

3 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago

An earthquake registered 7.4 on the Richter scale. If the reference intensity of this quake was 2.0 × 1011, what was its intensi

dexar [7]

32.0 you just add the first and last numbers then muptily the middle number by you anwser and divdie your 2 anwsers to gether

5 0

3 years ago

What is the square root of 1/23?

Helga [31]

0.04347826086

Step-by-step explanation:

this is the anwser,just round to whatever you need to

3 0

3 years ago

Read 2 more answers