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2 years ago

14

48 logs on 6 trucks

1 answer:

blagie [28]2 years ago

6 0

Answer:

288 is your answer. 8 is your answer.

Step-by-step explanation:

If you want to know how many logs all together, all you do is multiply.

48 x 6 = 288

288 is your answer.

Or division it is

48/6 = 8

8 is your answer.

You might be interested in

Which expression is equivalent to -3 1/8 minus 2 1/2

Sav [38]

-25/8 minus 5/2
(You just turn them into improper fractions)

8 0

2 years ago

Read 2 more answers

Edgar Anderson earns $300 per week plus 15% commission on all sales over 1,000. if Mr. Anderson's sales for one week are 2,500 w

vredina [299]

Since he only makes commission on sales over $1000, he makes commission on $1500 (we find this by doing 2500-1000.) His commission is 15%, so find 15% of $1500:
$.15*1500=225$
So we know how makes $225 in commission.

Now we need to add this to his fixed pay to get his total pay:
$225+300=525$
So his total pay is $525, which is answer 'a'

Hope I helped, and let me know if you have any questions :)

3 0

3 years ago

4. On a beach trip, Lucy rents a bike from "Wheels by the Waves", where

Dimas [21]

Answer:

6 hours

Step-by-step explanation:

The equation for this would be

c=3h+12

It costs $3 per hour, and there is a $12 cost no matter how many hours the bike is rented for

We know,c, or the cost is $30, so we can substitute that in

30=3h+12

Solve, by isolating the variable,h

First, subtract 12 on both sides

30=3h+12

18=3h

Divide by 3 on both sides

h=6

So, she rented the bike for 6 hours

7 0

2 years ago

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

2 years ago

Which two equations represent the statement “48 is 6 times as many as 8”

aleksandr82 [10.1K]

48=8•6
6•8=48

HOPE THIS HELPS

8 0

2 years ago