PLS HELP .........................

Graph the function represented in the table on the coordinate plane

yuradex [85]

Hello, the graph sould be like this. Just remember, if you want to draw a graph, you should know the x and y state. x is from left and y is from right.

4 0

3 years ago

Solve 1) 3x - 10 = 5x + 22 step by step please!!

emmainna [20.7K]

The answer is -16. Here's the explanation:
3 x - 10 = 5x + 22

3x - 5x = 22 + 10

-2x = 32

x = -16.
I hope this helped you!!!!!! :)

4 0

3 years ago

Read 2 more answers

The table shows the number of tickets each team sold. If they sold the same, how many did each sell? Remember to continue beyond

BartSMP [9]

Answer:

each sold 60 tickets

Step-by-step explanation:

Apparently, you're supposed to set the expressions equal and solve for t; then find the values of the expressions.

60(t -1) = 20(t +1)

60t -60 = 20t +20 . . . . . eliminate parentheses

40t = 80 . . . . . . . . . . . . . . add 60 -20t

t = 2 . . . . . . . . . . . . . . . . . . divide by 40

__

60(2 - 1) = 60 . . . . . solve for Falcons sales

20(2 + 1) = 60 . . . . .solve for Eagles sales (should be equal; it is)

Each team sold 60 tickets.

8 0

3 years ago

Heights of 10 year-olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6

IRISSAK [1]

Answer:

A normal probability plot of heights of a random sample of 500 10 year- old people should show a fairly straight line

Step-by-step explanation:

We are given that heights are normally distributed with mean=55 and standard deviation=6.

Now explaining all the given statements.

Roughly 95% of 10 year-old are between 37 and 73 inches tall.

This statement is not true. We know that approximately 95% of data lies within interval mean±2*standard deviation (empirical rule).

mean±2*standard deviation=55±2*6=55±12=(43,67)

So, 95% of 10 year-old are between 47 and 67 inches tall. Thus, the above statement is wrong.

A 10 year-old who is 65 inches tall would be considered more unusual than a 10 year-old who is 45 inches tall.

If the heights of 10 years old lie more than 2 standard deviation away then it will considered as unusual and If the heights of 10 years old lie more than 3 standard deviation away then it will considered as more unusual.

As 65 and 475 lies within two standard deviation from mean, so these are not unusual data values. So, the above statement is not true.

A normal probability plot of heights of a random sample of 500 10 year- old people should show a fairly straight line.

A normal probability plot shows straight line when the data is normally distributed and we know that if the population is normally distributed then then sample selected from this population is also normally distributed with mean μxbar and standard deviation σxbar. So, this statement is true.

We would expect more 10 year-old to be shorter than 55 inches than taller than 55 inches

Last few words were missing from the statement and i gathered them through web search.

The probability of 10 years old shorter than 55 is 50% and probability of 10 years old taller than 55 as area under the normal curve is 1 and given mean is 55. The area above mean and below mean in the normal curve is 0.5. So, we can't say that We would expect more 10 year-old to be shorter than 55 inches than taller than 55 as they have equal probabilities. Thus, the above statement is not true.

8 0

3 years ago

Help pls I'll give 25 points

RideAnS [48]

Answer:

$\text{C. }60$

Step-by-step explanation:

Question from image:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the x marks where it cannot be.

$\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}$

Let's start with the first position:

$\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in $3!=6$ ways. Therefore, there are $4\cdot 6=\boxed{24}$ possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

$\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in $3!=6$ ways. Therefore, there are $3\cdot 6=\boxed{18}$ possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are $3!=6$ ways to rearrange the remaining 3 digits for a total of $2\cdot 6=\boxed{12}$ possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still $3!=6$ ways to rearrange the remaining three numbers. Therefore, there are $1\cdot 6=\boxed{6}$ possible numbers when 1 is the fourth digit of the number.

Thus, there are $24+18+12+6=\boxed{60}$ numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

5 0

3 years ago