A clock on sale for $215 was regularly priced at $565 find the percent of discount

Just want to make sure that I did my maths right. anyone helps. thanks

ahrayia [7]

Yes you got them all correct at least I think.

5 0

3 years ago

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Plz help100 points. also, plz be the correct answer

vazorg [7]

For logarithmic equations,

log

b ( x ) = y

logb(x)=y

is equivalent to b y = x

by=x

such that x > 0

x>0 , b > 0

b>0 , and b ≠ 1

b≠1

.In this case, b = 5

b=5 , x = 25

x=25 , and y 2

y=2 .

b = 5

b=5

x =

25

x=25

y = 2

y=2

Substitute the values of b

b , x

x , and y

y

into the equation

b y =

x

by=x

.

5 squared

=

25

5squared=25

6 0

3 years ago

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How to do this question plz

bixtya [17]

252mmSquared

Step-by-step explanation:

28+28=56

74-56=18

18÷2=9

9×28=252

7 0

2 years ago

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The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

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