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2 years ago

suppose Q is the midpoint of PR. Use the information to find the missing value. PR = 9x-31 and QR=43; find x

Mathematics

1 answer:

In-s [12.5K]2 years ago

6 0

Answer:

x = 13

Step-by-step explanation:

If Q is the midpoint of PR...then

PR = 2QR...since QR is half the length of PR

;9x - 31 = 86

;9x = 86 + 31

;9x = 117

Hence x = 117/9 = 13

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A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= $\frac{6}{15} \times \frac{5}{14} = \frac{1}{7}$

Therefore the only statement that is not true is b.)

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2 years ago

Jennifer has a bag of chips that contains 2 red chips and 1 black chips. If she also has a fair die, what is the probability tha

Phoenix [80]

C) 16 bc if you think about it like 25% chance you going to pull the same chip out without looking

4 0

2 years ago

Read 2 more answers

What number should go in the space?<br> Multiplying by 1.15 is the same as increasing by _____%.

klasskru [66]

Answer:

115%

Because to get a percent from a decimal you multiply by 100 so you move the decimal twice to the right

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2 years ago

John wants to send a letter to peter who lives on tesla street john does not remember the house number however he knows that it

Serggg [28]

Answer: 137 letters.

Step-by-step explanation:

The data is:

The number has 4 digits.

The number is a multiple of 5

The number is a multiple of 7

The last digit is a 0.

Then, let's find all the possible numbers that met these conditions.

Our number can be formed with 4 digits, A, B, C and D.

D is the last one, so we have D = 0.

So the number is: ABC0

We know that all the numbers that end with 0 are multiples of 5, so that condition does not matter.

Now, a number is divisible by 7 if:

Take the last digit of the number, double it and subtract it to the remaining number, if the result is multiple of 7, then the initial number is a multiple of 7.

In this case the last digit is 0, so when we double it and subtract it, we actually are not changing the other 3 digits.

Then we have that:

ABC must be a multiple of 7.

So now our problem reduces to find the number of 3-digit numbers that are multiples of 7.

The first multiple of 7 that has 3 digits is:

7*15 = 105.

And the largest possible 3 digit number is 999, now let's try to divide it by 7 to see if this is also a multiple of 7.

999/7 = 142.7

the quotient is not integer, so 999 is not a multiple of 7.

The next option is 998, we can do the same here:

998/7 = 142.6

And so on, we will find that the largest 3 digit multiple of 7 is:

994, such that:

7*152=994.

Then we have:

Smallest: 7*15 = 105

Largest: 7*152 = 994.

Then the number of multiples of 7 between 999 and 100 will be equal to the difference between the quotients between the multples and 7.

105/7 = 15

994/7 = 152

152 - 15 = 137

So we have 137 multiples of 7 with 3 digits.

Then we have 137 4-digit numbers, that end with a zero and also are divisible by 5 and 7.

So John must send 137 letters if he wants to be shure that Peter will get the card.

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2 years ago

Simplify the expression: sin(7x/3)cos(5x/2) - cos(7x/3)sin(5x/2)

monitta

Ok no need to worry

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2 years ago