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3 years ago

15

Katie could already play five songs on her harp. She decided to learn to new songs each week. If S = songs and W = weeks, which

equation expresses her goals?

1 answer:

denis-greek [22]3 years ago

5 0

Answer:

she would have to record a the amount of songs She did every week

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If Jayna's shuffler is working properly and randomizing the playlist after each song, what is the probability of playing a slow

marta [7]

Answer:

1 song is a slow song, 2 songs are not

Step-by-step explanation:

that would give you 1/3 probability = 33.3333333333333%

4 0

3 years ago

Today, Olivia has $127 in her savings account. She will deposit $5 into the account each week. Given x>0 , which function rep

nevsk [136]

C because it's x means weeks so $5 a week is 5x and she already has 127 so 5x+127.

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3 years ago

A cylinder has a height of 10 inches and a radius of 5.5 inches. What is the

umka2103 [35]

Answer: 949.85 in 3

Step-by-step explanation:i took the test lol

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2 years ago

Lavania is studying the growth of a population of fruit flies in her laboratory. After 6 days she had nine more than five times

MatroZZZ [7]

Answer:

Lavania observed 39 fruit flies after 6 days of observation

Step-by-step explanation:

Let x be the number of fruit flies on the first day of Lavania's study.

After 6 days she had nine more than five times as many fruit flies as when she began the study.

Five times as many fruit flies as when she began the study = 5x

Nine more than five times as many fruit flies as when she began the study=5x+9

The expression to find the population of fruit flies Lavania observed after 6 days is 5x+9

If she observes 20 fruit flies on the first day of the study, then x=6, then

$5x+9=5\cdot 6+9=30+9=39$

5 0

3 years ago

The CEO of a large corporation asks his Human Resource (HR) director to study absenteeism among its executive-level managers at

Fynjy0 [20]

Answer:

Following are the answer to this question:

Step-by-step explanation:

Given:

n = 30 is the sample size.

The mean $\bar X$ = 7.3 days.

The standard deviation = 6.2 days.

df = n-1

$= 30-1 \\ =29$

The importance level is $\alpha$ = 0.10

The table value is calculated with a function excel 2010:

$= tinv (\ probility, \ freedom \ level) \\= tinv (0.10,29) \\ =1.699127\\ = t_{al(2x-1)}= 1.699127$

The method for calculating the trust interval of 90 percent for the true population means is:

Formula:

$\bar X - t_{al 2,x-1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar X+ t_{al 2,x-1} \frac{S}{\sqrt{n}}$

$=\bar X - t_{0.5, 29} \frac{6.2}{\sqrt{30}} \leq \mu \leq \bar X+ t_{0.5, 29} \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 \frac{6.2}{\sqrt{30}}\leq \mu \leq7.3 +1.699127 \frac{6.2}{\sqrt{30}}\\\\=7.3 -1.699127 (1.13196)\leq \mu \leq7.3 +1.699127 (1.13196) \\\\=5.37 \leq \mu \leq 9.22 \\$

It can rest assured that the true people needs that middle managers are unavailable from 5,37 to 9,23 during the years.

7 0

3 years ago