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3 years ago

I know the answer I just need a step by step explain. The answer is $1000

Mathematics

1 answer:

saveliy_v [14]3 years ago

6 0

sorry, I can help with this question but the question is cut off on the right, reply to me with the full question and id be happy to help :)

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Which decimal is greater 4.4 or 7.7

VARVARA [1.3K]

7.7 is the greater decimal

8 0

2 years ago

What expression does -5c=

hjlf

It equals 23 degrees Fahrenheit. So -5c=23f

5 0

3 years ago

If x is 1 what is the value of y?9x - 5y = 29

mylen [45]

Given

The equation,

9x - 5y = 29.

To find the value of y when x is 1.

Explanation:

It is given that,

$9x-5y=29$

That implies,

When x=1,

$\begin{gathered} 9(1)-5y=29 \\ 9-5y=29 \\ -5y=29-9 \\ -5y=20 \\ y=\frac{20}{-5} \\ y=-4 \end{gathered}$

Hence, the value of y is -4.

4 0

1 year ago

Necesito ayuda con está preguntas

prisoha [69]

Answer:

para el primer dibujo seria 1/12

para el segundo dibujo seria 1/16

para el tercer dibujo seria 1/10, pero tengo dudas con esta porque no se ve toda la figura.

Step-by-step explanation:

4 0

3 years ago

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding ma

IrinaVladis [17]

Answer:

538 books should be tested.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when $M = 0.08$

We have that $\sigma = 0.72$

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.08 = 2.575*\frac{0.72}{\sqrt{n}}$

$0.08\sqrt{n} = 2.575*0.72$

$\sqrt{n} = \frac{2.575*0.72}{0.08}$

$(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}$

$n = 537.1$

Rounding up

538 books should be tested.

6 0

2 years ago