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3 years ago

Please help !!!! Thank you

Mathematics

1 answer:

Jlenok [28]3 years ago

8 0

Answer:

Step-by-step explanation:

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Simplify (4x − 6) − (5x + 1). (1 point)

jok3333 [9.3K]

Answer:

-x -7

Step-by-step explanation:

(4x − 6) − (5x + 1)

Distribute the minus sign

(4x − 6) − 5x - 1

Combine like terms

4x-5x -6-1

-x -7

6 0

2 years ago

Scott wants to buy fencing to place around a semicircular rock garden The diameter of the semicircle is 16 feet Using 3.14 for p

Pavel [41]

The diameter of the semicircle is 16 feet.

He calculates that he needs to buy abut 25.12 feet of fencing.

To find amount of fencing we find the circumference of semicircle

Circumference of semicircle = $\frac{1}{2}\pi d+d$

Where d is the diameter of the circle

d= 16 given and value of pi is 3.14

Circumference of semicircle = $\frac{1}{2}*3.14*16+16$

=25.12 + 16 = 41.12 feet

Scott's calculation was wrong because he forgot to add the diameter d

Circumference of circle is half way of circle plus the diameter of circle.

Correct amount of fencing = 41.12 feet

7 0

2 years ago

A movie theater sells popcorn and fountain drinks. Brett buys 1 popcorn bucket and 3 fountain drinks for his family, and pays $9

siniylev [52]

Answer:

Step-by-step explanation:

p + 3d = 9.50

3p + 4d = 19.75

-3p - 9d = -28.50

3p + 4d = 19.75

-5d = -8.25

d= $1.65 drinks

p + 3(1.65) = 9.50

p + 4.95 = 9.50

p = $4.55 popcorn

7 0

2 years ago

Plzzzzzzxzz help with these 2 problems pleeeaseee:((

givi [52]

the first answer is 59.2 and the second answer is 1.5

4 0

2 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

2 years ago