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3 years ago

rachel is walking around her neighborhood. if she walks 1/3 mile in 1/6 hour , how fast in rachel walking in miles per hour?

1 answer:

4 0

2 miles per hour
all you have to do is multiply 1/3 by 6 since she only walks 1/3 miles in 1/6 of an hour and you want to find the miles walked in a hour

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Which method can't you use to solve this problem? x^2-47=0

Goryan [66]

Which method can't you use to solve this problem? x^2-47=0

☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆A.) Factoring
Cannot factor, no common factors.

B.) Squareroots

C.) Quadratic Formula

4 0

4 years ago

Read 2 more answers

A printer can print 27 pages in 4.5 min. How much time does it need to print 324 pages?

lutik1710 [3]

Answer:

54 minutes

Step-by-step explanation:

We can write a proportion to solve this problem. Put the number of pages over the minutes

27 pages 324 pages

-------------- = ---------------

4.5 minutes x minutes

Using cross products

27 * x = 324 * 4.5

27x = 1458

Divide each side by 27

27x/27 = 1458/27

x =54

6 0

4 years ago

Four different paints are advertised as having the same drying time. To check the manufacturer's claims, five samples were teste

Elina [12.6K]

Answer:

(1) 330

(2) 692

(3) 110

(4) 43.25

Step-by-step explanation:

The data provided is for the dying time of four different types of paint.

One-way ANOVA can be used to determine whether all the four paints have the same drying time.

Use Excel to perform the one-way ANOVA.

Go to Data → Data Analysis → Anova: Single Factor

A dialog box will open.

Select the data.

Select "Grouping" as Columns.

Press OK.

The output is attached below.

The required values are as follows:

(1)

Sum of Squares of Treatment (Between Subjects):

SST = 330

(2)

Sum of Squares of Error (Within Subjects):

SSE = 692

(3)

Mean Squares Treatment (Between Subjects):

MST = 110

(4)

Mean Squares Error (Within Subjects):

MSE = 43.25

4 0

3 years ago

TIME REMAINING

erik [133]

Answer:

$1,131.20 is the amount earned

Step-by-step explanation:

Key skills needed: Simple Interest Formula, Operations ( +, - , x , / )

1) To understand this problem, you need to know the simple interest formula.

A = P(1+rt)

A is the amount

P is the principal

R is the interest rate as a decimal

T is the time in years

2) The 1st thing we need to do is convert the interest rate into a decimal. We have 14%. To convert into decimal form, we divide it by a 100, or move the decimal 2 places to the left. This is 0.14 --> So r=0.14

3) Next we use the formula:

A = 1,010(1+ 0.14(8))

We first do 0.14(8) which is 1.12 and then add it to the 1 value, so you will get --> A = 1,010(2.12)
Multiply and you will get A = $2,141.20
To find the interest earned, you subtract this by the original amount, so $2,141.20 - $1,010 which would be $1,131.20

This means $1,131.20 is your answer.

Hope you understood and have a nice day!! :D

8 0

3 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

3 years ago