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1 year ago

12

Please please help ME!!!

1 answer:

Stella [2.4K]1 year ago

6 0

Answer:

720 Course Schedules

Step-by-step explanation:

When you choose the first course, there are 6 to choose from. When you scoose the second, there are only 5. This keeps going until the 6th course where there is only 1 choice.

You multiply these choices together. You should get 6×5×4×3×2×1

6×5=30

30×4=120

120×3=360

360×2=720

720×1=720

You then end up with 720 course schedules.

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3 years ago

A ball, thrown vertically upwards, from the ground, has its height h (in meters) expressed as a function of time t (in seconds),

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6 0

3 years ago

Read 2 more answers

Samantha has 4 1/2 gallons of lemonade for her daughter’s birthday party. Each gallon of lemonade contains 1.5 grams of fat. How

Lubov Fominskaja [6]

Answer:

6.75 grams

Step-by-step explanation:

Total amount of lemonade with Samantha = 4 1/2 gallons

= (4*2 + 1)/2 gallons = 9/2 gallons = 4.5 gallons

Amount of fat in 1 gallon of lemonade = 1.5 grams

To find amount of fat in 4.5 gallons lemonade

we multiply both side of equation with 4.5

Amount of fat in 1*4.5 gallon of lemonade = 1.5*4.5 grams = 6.75 grams

Thus,

there is 6.75 grams of fat in all the lemonade Samantha has for the party

3 0

2 years ago

I really need help on this 8th grade geometry

wariber [46]

The answer is 39 degrees <span />

5 0

2 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

2 years ago