What are the angles? HELP ASAP

A 2-digit number with x units and 7 less units than tens.

Hatshy [7]

Each two digit number has two numbers (duh!). Let's allow the tens digit to be x and the units digit to be y. Tens digit is 3 less than the units digit: x = y-3 Original number is 6 more than 4 times the sum of the digits: 10x+y-6 = 4x + 4y This gives us simultaneous equations!First let's clear the mess: 1. x= y-32. 6x-3y=6 Substitute 1 into 2: 6(y-3) -3y =66y - 18 - 3y = 6
3y = 24y = 8 Our units digit is 8 Substitute y= 8 into 1. x = y - 3x = 5 Our tens digit is 5 Therefore, our number is 58

7 0

3 years ago

The length of the segment indicated round your answer to the nearest 10th if nessesary

exis [7]

Pythagorean Theorem
x^2 = 15.7^2 -7.8^2
x^2 = 185.65 
x = 13.6253440324 
Answer is C

3 0

4 years ago

On an exam you get two points for every question answered correctly, zero points for each question left blank and lose one-half

Paraphin [41]

Using multiplication, the total score on the exam if you answered 13 questions correctly, leave 7 questions blank and answer 5 questions incorrectly is 39 point.

According to the question,

For each correct question, you get 2 points, and there are 22 correct questions, so multiply 22 by 2

22 * 2 = 44 points.

For each question left blank, you get 0 points, and there are 7 questions left blank, so multiply 7 by 0

7 * 0 = 0 points

For each incorrect question, you get -1 point, and there are 5 incorrect questions, so multiply 5 by -1

5 * (-1) = -5 points.

The total number of score = 22*2 + 7*0 + 5*(-1) = 44 + 0 - 5 = 39 points.

Hence, using multiplication, the total score on the exam if you answered 13 questions correctly, leave 7 questions blank and answer 5 questions incorrectly is 39 point.

Learn more about multiplication here

brainly.com/question/2865130

#SPJ4

3 0

2 years ago

The answer to a division problem is greater than the divisor always never or sometimes pick one

Ksenya-84 [330]

Sometimes.....
hope this helps! :)

8 0

3 years ago

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A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

quester [9]

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

6 0

3 years ago

Read 2 more answers