Please. I just need an answer. I don’t know how to mark brainliest but if I did I would.

Lina has 12 coins in her purse. D of them are dimes and the rest are quarters.

tester [92]

Answer:

we need more info but 11 q and 1 dime or 12 q and 0 dime

Step-by-step explanation:

Sorry need more info

7 0

3 years ago

Which answer represents the series in sigma notation?

Nat2105 [25]

The answer is b.......

8 0

2 years ago

Read 2 more answers

PLS ANSWER ASAP

Gnoma [55]

9514 1404 393

Explanation:

The inverse of a function y = f(x) can be found by solving x = f(y) for y.

When we do that here, we find ...

$x=f(y)\\\\x=\dfrac{y+a}{b}\\\\bx=y+a\\\\y=bx-a$

When we compare this to g(x), which we want to be the inverse of f(x), we see that ...

cx -d = bx -a ⇒ b=c, and d=a

We can choose a=d=1 and b=c=2 to make the two functions inverses:

f(x) = (x +1)/2

g(x) = 2x -1

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The inverse of f(x) is g(x) if f(g(x)) = x.

f(g(x)) = ((2x -1) +1)/2 = 2x/1 = x . . . . . g is the inverse of f

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Same as Part 2, but in reverse.

g(f(x)) = 2((x +1)/2) -1 = (x +1) -1 = x

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The attachment is a graph of the two functions and a table of values. The line y=x is the dashed orange line. You will notice that f(x) and g(x) are reflections of each other across that line.

6 0

3 years ago

a container of candy is shaped like a cylinder and has a volume of 125.6 cubic centimeters. If the heights of the container is 1

Elenna [48]

The radius of the container is 2 centimeter

<h3><u>Solution:</u></h3>

Given that a container of candy is shaped like a cylinder

Given that volume = 125.6 cubic centimeters

Height of conatiner = 10 centimeter

To find: radius of the container

We can use volume of cylinder formula and obatin the radius value

<em><u>The volume of cylinder is given as:</u></em>

$\text {volume of cylinder }=\pi r^{2} h$

Where "r" is the radius of cylinder

"h" is the height of cylinder and $\pi$ is constant has value 3.14

Substituting the values in formula, we get

$\begin{array}{l}{125.6=3.14 \times r^{2} \times 10} \\\\ {r^{2}=\frac{125.6}{31.4}} \\\\ {r^{2}=4}\end{array}$

Taking square root on both sides,

$r = \sqrt{4}\\\\r = 2$

Thus the radius of the container is 2 centimeter

4 0

3 years ago

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em> $\bar X$ <em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

<em>So, in the z table the P(Z </em> $\leq$ <em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

2 years ago