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3 years ago

5-3(n+4)-2.....please help

Mathematics

1 answer:

ELEN [110]3 years ago

4 0

Answer:

-3n-9

Step-by-step explanation:

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Use the distributive property to write an equivalent expression.<br> 7(1 – 10s + 8)

Ne4ueva [31]

Answer:

The answer is = −70s+63

Step-by-step explanation:

7(1−10s+8)

=(7)(1+−10s+8)

=(7)(1)+(7)(−10s)+(7)(8)

=7−70s+56

=−70s+63

5 0

3 years ago

Find the surface area of the cylinder. Use 3.14 for pi. Do not round the answer.

USPshnik [31]

Answer:

690.8

Step-by-step explanation:

I've done this one before

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2 years ago

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What is the area of this figure 3,7,7

IRINA_888 [86]

I think the answer is 147

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3 years ago

What does -2/3 (x + 5) + 4 (x + 5) - 10/3 (x + 5) equal

Kitty [74]

Answer: 0

Step-by-step explanation:

$-\frac{2}{3}(x+5)+4(x+5)-\frac{10}{3}(x+5) = ?$

$-\frac{2}{3}x-\frac{10}{3}x+4x+20-\frac{10}{3}x-\frac{50}{3} = ?$

$(-\frac{2}{3}x+4x -\frac{10}{3}x) +(-\frac{10}{3}+20-\frac{50}{3})=?$

$(-\frac{12}{3}x+4x)+(-\frac{60}{3}+20)=?\\ (-4x+4x)+(-20+20)=?\\0+0=0$

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3 years ago

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were

zloy xaker [14]

Answer:

The required probability is 0.927

Step-by-step explanation:

Consider the provided information.

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course.

That means 95% of students didn't enrolled in SAT prep course.

Let P(SAT) represents the enrolled in SAT prep course.

P(SAT)=0.05 and P(not SAT) = 0.95

30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students.

P(F) represents the first choice college.

The probability he didn't take an SAT prep course is:

$P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}$

Substitute the respective values.

$P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}$

$P[\text{not SAT} |P(F)]\approx0.927$

Hence, the required probability is 0.927

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3 years ago