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2 years ago

About how much would it have cost

Mathematics

2 answers:

Finger [1]2 years ago

6 0

One plant would cost about 0.50 cents.
You can find out the answer to this problem by dividing 8.15 by 16

Anton [14]2 years ago

4 0

It would cost about $.50 because $8.15 rounds to $8.00, and 8/16 = 0.5

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1. What percent of 58 is 18?

malfutka [58]

In order to find the answer you will have too divide 58 by 18 and then divide what you get from that by 100 to get your answer.

100%/x%=58/18

(100/x)*x=(58/18)*x

100=3.22222222222*x

(3.22222222222) to get x

100/3.22222222222=x

x=31.0344827586

Therefore your answer is "31%."

Hope this helps.

8 0

2 years ago

Solve 7k÷3=21<br><br><br> The solution is k __

qaws [65]

First, multiply by 3 on both sides:
$7k = 63$

Divide by 7:
$k = 9$

So, k = 9

4 0

2 years ago

Read 2 more answers

There are 24 girls on the track team. First race2/3 girl ran.only1/4 girl ran in second race. 1/6 on third race. Hiw many more g

alukav5142 [94]

Answer:

10 more girls

Step-by-step explanation:

So here 2/3 of 24 ran for first race

Then it gives us that 16 girls ran for first race.

Next, 1/4 of 24 ran for second race

That is 6 girls

The remaining girls ran for the third race,

24 - 22

2 girls

16 - 6 = 10

So for first race 10 more girls ran

5 0

2 years ago

Silicon wafers are scored and then broken into the many small microchips that will he mounted into circuits. Two breaking method

antoniya [11.8K]

Answer:

The estimate is $- 0.0265\le K \le 0.0465$

Method B this is because the faulty breaks are less

Step-by-step explanation:

The number of microchips broken in method A is $n_1 = 400$

The number of faulty breaks of method A is $X_1 = 32$

The number of microchips broken in method B is $n_2 = 400$

The number of faulty breaks of method A is $X_2 = 32$

The proportion of the faulty breaks to the total breaks in method A is

$p_1 = \frac{32}{400}$

$p_1 = 0.08$

The proportion of the faulty to the total breaks in method B is

$p_2 = \frac{28}{400}$

$p_2 = 0.07$

For this estimation the standard error is

$SE = \sqrt{ \frac{p_1 (1 - p_1)}{n_1} + \frac{p_2 (1- p_2 )}{n_2} } }$

substituting values

$SE = \sqrt{ \frac{0.08 (1 - 0.08)}{400} + \frac{0.07 (1- 0.07 )}{400} } }$

$SE = 0.0186$

The z-values of confidence coefficient of 0.95 from the z-table is

$z_{0.95} = 1.96$

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as

$K = [p_1 - p_2 ] \pm z_{0.95} * SE$

substituting values

$K = [0.08 - 0.07 ] \pm 1.96 *0.0186$

$K = - 0.0265 \ or \ K = 0.0465$

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is

$- 0.0265\le K \le 0.0465$

3 0

2 years ago

What is the square root of 56 in radical form

V125BC [204]

$\sqrt{56} \\ \\ \sqrt{7*8} \\ \\ \sqrt{7*4*2} \\ \\ \sqrt{7*2*2*2} \\ \\ 2 \sqrt{7*2} \\ \\ 2 \sqrt{14}$

7 0

2 years ago