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2 years ago

Dolores bought 15 party hats priced at $0.75 each and 15 noisemakers priced at $1.25 each. How much did Dolores spend in all?

Mathematics

2 answers:

Ludmilka [50]2 years ago

8 0

She spends 30 dollars

Naddik [55]2 years ago

4 0

The answer is 30$!!

i hope this helps and have a wonderful day!

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Amy is just learning how to rock climb. Her instructor takes her to a 26 ft climbing wall for her first time. She climbs up 5 ft

boyakko [2]

Let's compute the speeds as she goes up and down of the climbing wall. Speed is the ratio of distance to time.

Speed going up = 5 ft/(2min * 60 s/1 min) = 1/24 ft/s
Speed going down = 2 ft/10 s = 0.2 ft/s
Net speed = 1/24 ft/s - 0.2 ft/s = 5/24 ft/s

Using this net speed, we can already calculate for the total time:

Speed = Distance/Time
5/24 ft/s = 26 ft/Time
Time = 124.8 seconds or 2.08 minutes

7 0

2 years ago

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

2 years ago

Mr. Peterson needs topsoil for his garden. His rectangular garden is 78 in long and 48 in wide. A bag of topsoil covers an area

nata0808 [166]

Answer:

3744 inches squared, and 8 bags

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

˜”*°•.˜”*°• Area: •°*”˜.•°*”˜

Well, the formula for area is L*W

L=78 inches

W=48 inches

78*48=3744 inches squared

˜”*°•.˜”*°• Number of bags: •°*”˜.•°*”˜

3744/500=7.448

So he would need to buy 8 bags

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

4 0

1 year ago

Factor completely. <br> 81x^2+180x+100

Tasya [4]

(9x+10)² or (9x+10)(9x+10)

8 0

2 years ago

1. A researcher is interested in studying whether or not listening to music while jogging makes people run faster. He thinks tha

castortr0y [4]

Step-by-step explanation:

1.Assuming the same sample size and considering the same value for the errors ( not taking into consideration the type of music, the volume of the sound and cow familiar the runner is with that type of stimuli, age group, time of the day/ number of days, running conditions like wether and equipment, distance) one can state:

A. Music has no influence over the running speed ( when jogging) in Washington DC

B.When listening to music, people ( in Washington DC) run faster while jogging

The mean running speed is a simple, ponderate or other type of mean ( that takes into consideration the variations of speed at the beginning and by the end of the race?

4 0

2 years ago