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I've been on this question for 15 min..... please help

Elanso [62]

I believe the answer would be no

7 0

3 years ago

Read 2 more answers

Barry found a computer at Staples for $989. A week later, the computer went on sale for $849. What

JulijaS [17]

Barry found a computer at Staples for $989. A week later, the computer went on sale for $849. The percentage change is 14.156 % decrease

Solution:

Given that Barry found a computer at Staples for $989

A week later, the computer went on sale for $849

To find: percent of change

Percent change is the extent to which a variable gains or loses value. The figures are arrived at by comparing the initial (or before) and final (or after) quantities according to a specific formula.

THE PERCENT CHANGE IS GIVEN AS:

$\text{ percent change } =\frac{\text { final value - initial value }}{\text { initial value }} \times 100$

If the result is positive, it is percentage increase

If the result is negative, it is percentage decrease

Here initial value = 989 and final value = 849

Substituting the values in above formula,

$\begin{aligned}&\text { percent change }=\frac{849-989}{989} \times 100\\\\&\text { percent change }=-0.14155 \times 100=-14.156 \%\end{aligned}$

Here negative sign denotes percentage decrease

Thus percentage change is 14.156 % decrease

8 0

3 years ago

On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas,

erastova [34]

Answer:

a

The distribution of X is normal

b

$P(X < 1.4) = 0.12654$

c

$P(X > 3.5) = 0.36051$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 3 \ hours$

The standard deviation is $\sigma = 1.4 \ hours$

Generally given from the question that the amount of time spent alone by the population size is normally distributed then then the distribution of X (i.e the amount of time spent by the sample size (the one Mercurian)) will be normally distributed

Generally the probability that the child spend less than one hour in a day is mathematically represented as

$P(X < 1.4) = P(\frac{X - \mu}{\sigma} < \frac{1.4 - \mu}{\sigma} )$