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3 years ago

I need help with this problem 23 in to 61,982 not sure how to work the problem

Mathematics

1 answer:

Korolek [52]3 years ago

6 0

What exactly is the problem? , there isn’t a photo or a description of what the problem is.

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Solve using factoring

irina1246 [14]

Answer: you can use Photomath

Step-by-step explanation:

4 0

3 years ago

Latisha worked for six hours and 45 minutes how should she write that on her time card

AlexFokin [52]

Answer:

Hey there!

6 hrs 45 mins = 405 mins, or 6.75 hours.

Hope this helps :)

8 0

3 years ago

A model satellite has a scale of 1 in: 6 ft.

insens350 [35]

Answer:

B) 18 ft

Step-by-step explanation:

The scale is 1 in. model : 6 ft real.

6/1 = 6

To go from model size to real size, multiply the model size by 6 and change from inches to feet.

3 in. model = 3 × 6 ft = 18 ft

5 0

1 year 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

There are 13 boys and 16 girls in Ms. Lau's class. What is the ratio of boys to the whole class?

saveliy_v [14]

Answer:

13:29

Step-by-step explanation:

16+13= 29

4 0

3 years ago

Read 2 more answers