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2 years ago

Simplify 1.5m7(-4m5)2

Mathematics

1 answer:

ANTONII [103]2 years ago

3 0

ANSWER: -12m12

I’m positive I’m correct, Good luck!

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A bowler knocks at least 6 pins 70% of the time.Out of 200 rolls, how many times can you predict the bowler will knock down at l

ankoles [38]

The bowler will hit 6 pins 140 times because 70% of 200 is 140

6 0

3 years ago

Christine is 4 years younger than Mark. 3 years from now, she will be two-thirds as old as Mark find their present ages.Complete

icang [17]

Answer:Mark Is 9 years old; Christine is 5 years old

Step-by-step explanation:

Present age

LET mark age be x

Christine be x-4 yrs

In 3 yrs Time

Mark will be x+3 yrs old

Christine will be (x -4 +3) = x-1 years old

Also, In 3 years time

Christine = 2/3 Mark

such that

x-1 = 2/3(x+3)

x -1= 2x+6 / 3

3x -3= 2x+6

3X--2X = 6+3

=X=9

Mark Is 9 years old

Christine is x -4 = 9-4=5 years old

5 0

2 years ago

Make w the subject of the formula s = w/a

enot [183]

Answer:

w = sa

Step-by-step explanation:

Given

s = $\frac{w}{a}$ ( multiply both sides by a to clear the fraction )

sa = w

8 0

2 years ago

John has 12 pounds of dog food and is going to separate it into 3/4 pound portions. How many portions of dog food will he have?

NNADVOKAT [17]

16 portions

explanation:

12 / 3/4

= 12 * 4/3

= 4*4 = 16

7 0

2 years ago

The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s

lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

$MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}$

Rearranging this equation in terms of n gives

$n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2$

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

$n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n$

As n is given as 40 so the new sample size is given as

$n_2=4n\\n_2=4*40\\n_2=160$

So the sample size to obtain the desired margin of error is 160.

4 0

3 years ago