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3 years ago

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A pair of jeans that normally sells for $35 is in sale for 20%. Find the sale price of the jeans. Then find the total cost of th

e jeans if the sales tax rate is 6%

1 answer:

never [62]3 years ago

7 0

35×.2= 7 dollars is being taken away from the sale.

35-7= 28 dollars which is the new cost.

28×.06=1.68
28+1.68=29.68 final cost of jeans

or you can do 28×1.06= 29.68 final cost of jeans.

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Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Pls answer show work, At a store, notebooks are on sale for 78% off the regular price. The cost of a notebook is represented by

Assoli18 [71]

Answer:

<h2>C</h2>

Step-by-step explanation:

<h3> dont subtract the cost of the notebook by 0.78 while multiping it with the cost.</h3>

5 0

3 years ago

sergij07 [2.7K]

Re read the question again just in case, try and put £25.50
But I got 25.5 too

4 0

2 years ago

Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636, the central limit theorem appl

lakkis [162]

Answer:

a) $\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)$

$\mu_{\bar X}=100$ $\sigma^2_{\bar X}=1$

b) $P(\bar X >101)=1-P(\bar X$

c) $P(\bar X$

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".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

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

$X \sim N(\mu=100,\sigma=6)$

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

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

a. What are the mean and variance of the sampling distribution for the sample means?

$\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)$

$\mu_{\bar X}=100$ $\sigma^2_{\bar X}=1.2^2=1.44$

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833$

And we want to find this probability:

$P(\bar X >101)=1-P(\bar X$

On this last step we use the complement rule.

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

$z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67$

And we want to find this probability:

$P(\bar X$

5 0

3 years ago

I don't really understand this??....

Mila [183]

Answer:

Part 1

its asking you to find the slope of the given lines by using the points that they have plotted on the lines

first you need to figure out what the points are, then insert it into the formula

the formula for this would be $\frac{Y2- Y1}{X2-X1}$

Part 2:

its asking you to find the slope using the points that are written in each box

again, just substitute it into the formula

the formula for this would be $\frac{Y2- Y1}{X2-X1}$

5 0

2 years ago