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2 years ago

6

The zoom feature on a copier sets the percent size of a copy. Carol wants to make a 150% copy of a photograph. The original phot

o has a width of 8 inches. What will be the width of the copy?
A. 5.3 in.
B. 12 in.
C. 25 in.
D. 120 in.

i need help ASAP

2 answers:

Nookie1986 [14]2 years ago

7 0

Answer:

1.5

Step-by-step explanation:

Vsevolod [243]2 years ago

3 0

150% = 1.5

multiply 8 by 1.5

8 * 1.5 = 12

so the answer is B

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17) Solve the system as matrix equations using inverses. 17) A company produces three models of MP3 players, models A, B, and C.

Vinvika [58]

Answer:

Model A=30, Model B: 20, Model C: 45

Step-by-step explanation:

Let x be the amount of type A mp3 produced, and the amount of model B mp3's produced and z the amount of model C mp3 produced.

Since in total there are 303 hours for electronics work, then:

$3.2x+5.4y+2.2z=303$

Since in total there are 393 hours for assembly, then:

$2.8x+2.4y+5.8z=393$

Since in total there are 416 hours for quality assurance, then:

$4.4x+3.4y+4.8z=416$

Then, the linear system associated to the problem is

$3.2x+5.4y+2.2z=303\\2.8x+2.4y+5.8z=393\\4.4x+3.4y+4.8z=416$

with coefficient matrix $A=\left[\begin{array}{ccc}3.2&5.4&2.2\\2.8&2.4&5.8\\4.4&3.4&4.8\end{array}\right]$ and vector of constant terms $b=\left[\begin{array}{ccc}303\\393\\416\end{array}\right]$

Since the determinant of A is equal to 36.704 then A is invertible.

Then for solve the system $Ax=b$ , is enough find the inverse of A and operate

$Ax=b\\A^{-1}Ax=A^{-1}b\\x=A^{-1}b$

Using Octave we obtain that

$A^{-1}=\left[\begin{array}{ccc}-0.22&-0.50 &0.71\\0.33&0.15&-0.34\\-0.028& 0.35&-0.20\end{array}\right]$

Then

$x=A^{-1}b=\left[\begin{array}{ccc}30\\20\\45\end{array}\right]$

This means that 30 mp3's of model A, 20 of model B and 45 of model C must be produced

8 0

3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that $\mu = 50$ and $\sigma = 6$ .

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of $X = 60$ is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

$X = 67$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

$p = .9977 - 0.9525 = .0452$

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

$Z = \frac{X - \mu}{\sigma}$

$-1.29 = \frac{X - 50}{6}$

$X - 50 = -7.74$

$X = 42.26$

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 7.68$

$X = 57.68$

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0

3 years ago

Can someone please help me? I will give thanks!!

defon

I'm pretty sure it would be d
Hope this helps!

7 0

3 years ago

Read 2 more answers

I will mark brainliest for correct answer

statuscvo [17]

24, it’s the lcf and is the closest number that they have in common

3 0

2 years ago

Read 2 more answers

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NeTakaya

The answer is going to be B

5 0

2 years ago