Which statements are true select each correct answer

How much of a spice that is 30% salt should be added to 175 ounces of a spice that is 65% salt in order to make a spice that is

klasskru [66]

$\bf \begin{array}{lccclll}
&amount&concentration&
\begin{array}{llll}
concentrated\\
amount
\end{array}\\
&-----&-------&-------\\
\textit{30\% spice}&x&0.30&0.30x\\
\textit{65\% spice}&175&0.65&(175)(0.65)\\
-----&-----&-------&-------\\
mixture&x+175&0.50&(x+175)(0.50)
\end{array}$

so.. as you can see, we use the decimal notation for the percentages... 65% is just 65/100 and 50% is just 50/100 and so on

so.. whatever the salted amounts are, we know they'll add up to (x+175)(0.50)

thus (0.30x) + (175)(0.65) = (x+175)(0.50)

solve for "x"

5 0

3 years ago

A bag contains 5 back marbles 10 red marbles and 20 blue marble which statement is true

Galina-37 [17]

What are the statements?

4 0

2 years ago

The mean of a set of data is 148.87 and its standard deviation is 68.29. Find the z score for a value of 490.19.

Ksivusya [100]

The mean of a set of data is 148.87 and its standard deviation is 68.29. Find the z score for a value of 490.19
the z-score is given by:
z=(x-μ)/σ
plugging in the values in the expression we get:
z=(490.19-149.87)/68.29
z=340.32/68.29
z=4.9835

6 0

3 years ago

Jack can build 1/5 of a shed in the same time Kyle build 5/8 of a shed. How much of a shed will jack have built when Kyle has fi

nadya68 [22]

Answer:

Step-by-step explanation:

If Kyle can get 5/8 of a shed built in x amount of time, he can get 62.5% done; if Jack can get 1/5 of the shed built in the same time, he can 20% of it done. As a ratio:

$\frac{Kyle}{Jack}:\frac{62.5}{20}$ Kyle gets the whole thing done, 100% of the way, leaving Jack lagging behind at a somewhat lower percentage of the work being done.

$\frac{Kyle}{Jack}:\frac{62.5}{20}=\frac{100}{x}$ and cross multiply to solve for the percentage of the shed built by Jack by the time Kyle gets it completed:

62.5x = 2000 so

x = 32%

4 0

3 years ago

With a height of 68 in, Nelson was the shortest president of a particular club in the past century. The club presidents of the

Ivahew [28]

Answer:

a. The positive difference between Nelson's height and the population mean is: $\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: $\\ z = -1.1739 \approx -1.174$ (Nelson's height is below the population's mean 1.174 standard deviations units).

d. Nelson's height is usual since $\\ -2 < -1.174 < 2$ .

Step-by-step explanation:

The key concept to answer this question is the z-score. A z-score "tells us" the distance from the population's mean of a raw score in standard deviation units. A positive value for a z-score indicates that the raw score is above the population mean, whereas a negative value tells us that the raw score is below the population mean. The formula to obtain this z-score is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

$\\ z$ is the z-score.

$\\ \mu$ is the population mean.

$\\ \sigma$ is the population standard deviation.

From the question, we have that:

Nelson's height is 68 in. In this case, the raw score is 68 in $\\ x = 68$ in.
$\\ \mu = 70.7$ in.
$\\ \sigma = 2.3$ in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson's height and the mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

$\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

That is, the positive difference is 2.7 in.

b. How many standard deviations is that [the difference found in part (a)]?

To find how many standard deviations is that, we need to divide that difference by the population standard deviation. That is:

$\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174$

In words, the difference found in part (a) is 1.174 standard deviations from the mean. Notice that we are not taking into account here if the raw score, x, is below or above the mean.

c. Convert Nelson's height to a z score.

Using formula [1], we have

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

$\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}$

$\\ z = \frac{-2.7\;in}{2.3\;in}$

$\\ z = -1.1739 \approx -1.174$

This z-score "tells us" that Nelson's height is 1.174 standard deviations below the population mean (notice the negative symbol in the above result), i.e., Nelson's height is below the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider "usual" heights to be those that convert to z scores between minus2 and 2, is Nelson's height usual or unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

$\\ -2 < z_{Nelson} < 2$

$\\ -2 < -1.174 < 2$

Then, Nelson's height is usual according to that statement.

7 0

3 years ago

Which statements are true select each correct answer​

Which statements are true select each correct answer