Draw an area model. Then, Solve using the standard algorithm. 642 x 207

5. Determine the value of every variable in the rhombus below.

Zina [86]

Answer:

m = 160

p = 90

w = 10

z = 80

y = 10

Step-by-step explanation:

7 0

2 years ago

Help me solve -2+5n=4n+2

iris [78.8K]

Sorry for the messy handwriting

7 0

3 years ago

The time taken to deliver a pizza has a uniform probability distribution from 20 minutes to 60 minutes. What is the probability

Whitepunk [10]

Answer:

(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2a) The percentage of results more than 45 is 79.67%.

(2b) The percentage of results less than 85 is 91.77%.

(2c) The percentage of results are between 75 and 90 is 15.58%.

(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

Step-by-step explanation:

(1)

Let Y = the time taken to deliver a pizza.

The random variable Y follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

$f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.$

Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:

$P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70$

Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2)

Let X = results of a certain blood test.

It is provided that the random variable X follows a Normal distribution with parameters $\mu = 60$ and $s = 18$ .

The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.

The z-scores follows a Standard normal distribution, N (0, 1).

(a)

Compute the probability that the results are more than 45 as follows:

$P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z$

The percentage of results more than 45 is: $0.7967\times100=79.67\%$

Thus, the percentage of results more than 45 is 79.67%.

(b)

Compute the probability that the results are less than 85 as follows:

$P(X$

The percentage of results less than 85 is: $0.9177\times100=91.77\%$

Thus, the percentage of results less than 85 is 91.77%.

(c)

Compute the probability that the results are between 75 and 90 as follows:

$P(75$

The percentage of results are between 75 and 90 is: $0.1558\times100=15.58\%$

Thus, the percentage of results are between 75 and 90 is 15.58%.

(d)

Compute the probability that the results are between 20 and 100 as follows:

$P(20$

Then the probability that the results outside the range 20 to 100 is: $1-0.9736=0.0264$ .

The percentage of results outside the range 20 to 100 is: $0.0264\times100=2.64\%$

Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.

4 0

3 years ago

Substitution of y= 4x-3 3x-2y=16

bearhunter [10]

Answer:

(-2,-11)

Step-by-step explanation:

3x-2(4x-3)=16

3x-8x+6=16

-5x=10

x=-2

Plug back in

y=4(-2)-3

y=-8-3

y=-11

4 0

3 years ago

Read 2 more answers

A survey found that 12 out of 30 people preferred bacon over sausage. According to this survey, what percent of people prefer ba

postnew [5]

Given

A survey found that 12 out of 30 people preferred bacon over sausage.

Find out the what percent of people prefer bacon over sausage.

To proof

As given in the question

survey found that 12 out of 30 people preferred bacon over sausage.

Total number of people = 30

people preferred bacon over sausage = 12

Formula

$Percentage =\frac{Part}{Total} \times 100$

put the value in the formula

$Percentage\ of\ prefer\ bacon\ over\ sausage = \frac{12}{30} \times 100$

solve the above

percent of people prefer bacon over sausage = 40 %

Hence proved

3 0

3 years ago