Need help asap.(Plz answer this)

Nastasia [14]

Answer:

54

Step-by-step explanation:

$\frac{1}{3x} - \frac{1}{2} = 18 \frac{1}{2} \\ \\ \frac{1}{3x} - \frac{1}{2} = \frac{37}{2} \\ \\ \frac{1}{3x} = \frac{37}{2} - \frac{1}{2} \\ \\ \frac{1}{3x} = \frac{37 - 1}{2} \\ \\ \frac{1}{3x} = \frac{36}{2} \\ \\ \frac{1}{3x} = 18 \\ \\ 1 = 54x$

8 0

3 years ago

Whoever gets this correct gets brain, 5 stars and a thanks! An explanation of it would be amazing! Tysm!

vlabodo [156]

Answer:

C and E and F

Step-by-step explanation:

Integers are basically whole numbers. For example, 6.5 is not an integer since it is not a whole number and it is between 6 and 7. Also fractions that cannot be simplified to whole numbers are not integers. Also integers can be negative.

6 0

3 years ago

A chemist has three different acid solutions,

serious [3.7K]

9514 1404 393

Answer:

36 liters of 20%
18 liters of 35%
54 liters of 55%

Step-by-step explanation:

Let a, b, c represent the quantities of 20%, 35%, and 55% solutions, respectively. Then for the given mix, we have ...

a + b + c = 108 . . . . . total number of liters

.20a +.35b +.55c = 0.40·108 . . . . . liters of acid

0a - 3b + c = 0 . . . . . 55% solution was 3 times 35% solution

__

Solving these three equations by your favorite method gives ...

(a, b, c) = (36, 18, 54)

The chemist used 36 liters of 20%, 18 liters of 35%, and 54 liters of 55%.

_____

<em>Comment on alternate approach</em>

Since the ratio of 55% to 35% is 3 : 1, that mix will have an acid content of ...

(3(.55) +1(.35))/4 = 0.50 = 50%

So, the final mix is equivalent to some quantity of 50% mix being added to some quantity of 20% mix. The fraction that is 50% mix* will be (40-20)/(50-20) = 2/3 of the total, or (2/3)(108 L) = 72 L.

Now we know that (1/3)(108 L) = 36 L of 20% solution is needed and (1/4)(72 L) = 18 L of 35% solution is needed. 3 times that is 54 L of 55% solution.

__

* After you work a few mixture problems, you can see the pattern. If we start with an equation for x = fraction of 50% to use with 20% to make 40%, we get ...

x(50%) +(1 -x)(20%) = 40%

x(50% -20%) = 40% -20% . . . . . subtract 1(20%) and factor out x

x = (40% -20%)/(50% -20%)

The key here is to see where the numbers 20%, 40%, and 50% show up in the fraction.

3 0

4 years ago

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

4 years ago

Read the following passage and answer the question.

olga2289 [7]

Answer:

The Correct answer is:

A. You aren’t telling me anything I don’t already know.

Step-by-step explanation:

4 0

3 years ago

Need help asap.(Plz answer this)​

Need help asap.(Plz answer this)