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2 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Mathematics

1 answer:

Tanzania [10]2 years ago

6 0

Answer:

Approximately, 159 men weighs more than 165 pounds and 159 men weighs less than 135 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 pounds

Standard Deviation, σ = 15

We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P( men weighing more than 165 pounds)

P(x > 165)

$P( x > 165) = P( z > \displaystyle\frac{165 - 150}{15}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 165) = 1 - 0.8413 = 0.1587 = 15.87\%$

Approximately, 159 men weighs more than 165 pounds.

P(men weighing less than 135 pounds)

P(x < 135)

$P( x < 135) = P( z < \displaystyle\frac{135 - 150}{15}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 135) = 0.1587 = 15.87\%$

Approximately, 159 men weighs less than 135 pounds.

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For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How

natulia [17]

<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let $e$ the cost of the exam hours

Let $w$ be the cost of the workshop hours

Let $l$ be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

$e=2w$ equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

$w=2l$ equation (2)

Finally, we also know that 3hr exams 24hr workshops and 12hr lectures cost $528, so:

$3e+24w+12l=528$ equation (1)

Now, lets find the value of $l$ :

Step 1. Solve for $l$ in equation (3)

$3e+24w+12l=528$

$12l=528-3e-24w$ equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

$12l=528-3e-24w$

$12l=528-3(2w)-24w$

$12l=528-6w-24w$

$12l=528-30w$ equation (5)

Step 3. Replace equation (2) in equation (5) and solve for $l$

$12l=528-30w$

$12l=528-30(2l)$

$12l=528-60l$

$72l=528$

$l=\frac{528}{72}$

$l=\frac{22}{3}$

$l=7.33$

Cost of lectures = $7.33 per hour

3 0

2 years ago

Read 2 more answers

A farm can harvest 120 pounds of carrots per acre of land. Each crate can hold 24 pounds of carrots. If a farmer owns 4.8 acres

Shtirlitz [24]

Answer:

The farmer needs 24 crates to hold the carrots.

Step-by-step explanation:

To determine how many crates the farmer needs to hold the carrots, knowing that his farm has 4.8 acres of land, where he can harvest 120 pounds per acre, and that each crates can hold up to 24 pounds of carrots, the following calculation must be performed:

(4.8 x 120) / 24 = X

576/24 = X

24 = X

Thus, the farmer needs 24 crates to hold the carrots.

6 0

1 year ago

For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a

aleksley [76]

Answer:

a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

Option c) .253 is.correct

c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

= 0.344

Option c is.correct

d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

7 0

2 years ago

The prices of three t-shirts styles are $24, $30 and $36. the probability of choosing a $24 t-shirt is 1/6. the probability of c

Slav-nsk [51]

$\text{Answer} : \text{The expected value of a t-shirt is \$31.}$

Explanation:

Since we have given that

The prices of three t-shirts styles i.e $24, $30, $36 with their probability is given by

$\frac{1}{6}, \frac{1}{2},\frac{1}{3}$

As we know that,

$E(X)= \sum_{1}^{3}x_iP(x_i)$

$\text{where} x_i \text{ is the prices of t- shirts styles}$

Now,

$x_1= \$24 , x_2=\$30 , x_3=$36$

and

$P(x_1)=\frac{1}{6},P(x_2)=\frac{1}{2}, P(x_3)=\frac{1}{3}$

So,

$E(X)= 24\times \frac{1}{6}+30\times\frac{1}{2}+36\times \frac{1}{3}\\=4+15+12\\=31$

So, the expected value of a t-shirt = $31.

4 0

2 years ago

A weatherman collected data on snow accumulation. A line of best fit was computed. The equation for the line is: y = 1.5x + 0.12

SVETLANKA909090 [29]

Answer:

5x-12=42 over 3 subtract that and you will get 1

Step-by-step explanation:

8 0

2 years ago