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2 years ago

Find the percent of change. Round to the nearest percent. Original: 9.8 New: 12.1

Mathematics

2 answers:

umka2103 [35]2 years ago

8 0

so my calculations show that (12.1 / 9.8) - 1 =~ 0.2346938775510204081632653061224 that is about a percentage of 23.469% thank you for asking that question, oh and this answer is also on yahoo

andrew11 [14]2 years ago

6 0

Answer:

The percent of change is 23% ( approx )

Step-by-step explanation:

Given,

The original value = 9.8,

New value = 12.1

Thus,

$\text{The percentage change}=\frac{\text{New value-original value}}{\text{Original value}}\times 100$

$=\frac{12.1-9.8}{9.8}\times 100$

$=\frac{230}{9.8}$

$=23.4693877551\%$

$\approx 23\%$

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Carlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has. Wh

notka56 [123]

Answer:

<h2>He could use the medium box.</h2>

Step-by-step explanation:

Notice that the umbrella is 37.5 inches long.

If you observe, each box doesn't have 37 inches wide, however, they are tridimensional figures, which means they have a certaing height.

So, the smalle box is too small to fit the umbrella.

But, the medium box is perfect, because it has 27 inches wide and 27 inches height, which is enough to fit the umbrella. These are the best dimensions to send the umbrella.

Therefore, the medium box is the answer, the one that has dimensions 27x14x27.

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2 years ago

Read 2 more answers

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.

In-s [12.5K]

Answer:

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=15,\sigma=4)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case $\bar X \sim N(15,\frac{4}{\sqrt{64}})$

Solution to the problem

We are interested on this probability

$P(13.5$

If we apply the Z score formula to our probability we got this:

$P(13.5$

$=P(\frac{13.5-15}{\frac{4}{\sqrt{64}}}$

And we can find this probability on this way:

$P(-3$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-3$

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.

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2 years ago

If VX = WZ = 40 cm and m∠ZVX = m∠XWZ = 22°, can ΔVZX and ΔWXZ be proven congruent by SAS? Why or why not?

Paul [167]

Answer: The ΔVZX and ΔWXZ are not congruent by SAS.

Explanation:

It is given that the VX = WZ = 40 cm and ∠ZVX = ∠XWZ = 22°.

Draw a figure as shown below,

According to the SAS rule of congruence, two triangles are congruent if two sides and their inclined angle is equal.

From the given figure it is easily noticed that in ΔVZX and ΔWXZ,

$VX=WZ=40cm$ (given)

$\angle ZVX=\angle XWZ=22^{\circ}$ (given)

$XZ=XZ$ (common side)

Since we have two sides and one angle is same. but we can not conclude that the ΔVZX and ΔWXZ are congruent by SAS, because the given angle is not the inclined angle of both equal sides.

Therefore, the ΔVZX and ΔWXZ are not congruent by SAS.

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2 years ago

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Lina20 [59]

-3(y+2)^2-5+6y
-3(y^2+4y+4)-5+6y
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2 years ago

Hotels’ use of ecolabels. Refer to the Journal of Vacation Marketing (January 2016) study of travelers’ familiarity with ecolabe

Alexxx [7]

Answer:

a) 0.7486; b) 0.6518; c) No

Step-by-step explanation:

We use z scores to answer these questions. The formula for a z score is

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

Our mean, μ, is 44; our standard deviation, σ, is 1.5.

For part a,

We want P(X > 43):

z = (43-44)/1.5 = -1/1.5 = -0.67

Using a z table, we see that the area under the curve to the left of this is 0.2514. However, we want the area to the right; this means we subtract from 1:

1 - 0.2514 = 0.7486

For part b,

We want P(42 ≤ X ≤ 45):

z = (42-44)/1.5 = -2/1.5 = -1.3

z = (45-44)/1.5 = 1/1.5 = 0.67

Using a z table, the area under the curve to the left of z = -1.3 is 0.0968. The area under the curve to the left of z = 0.67 is 0.7486.

The area between them is 0.7486 - 0.0968 = 0.6518.

For part c,

A response of 35 has a z score of

z = (35-44)/1.5 = -9/1.5 = -6

This is 6 standard deviations below the mean. This is more than the amount required to make this an outlier, or an unlikely response.

8 0

2 years ago