All categories

3 years ago

Will mark brainiest

2 answers:

5 0

Ime to walk uphill : t1 = d/3
time to walk downhill: t2 = d/6
total distance = 2d ( up hill and down ill)
2d = (d/3 + d/6)r
2/(1/3 +1/6) = r
r = 4
average speed up and down the hill was 4 mph

total distance = (2/3 x 6 x 4) + (1/3 x 6 x 4)
total distance = 16 + 8 = 24 miles total

Average speed to top of hill = 2/(1/4 +1/3) = 3.43 mph

to top of hill = 24/2 = 12 miles

12 / 3.43 = 3.5 hours
12 oclock + 3.5 hours = 3:30

total miles was 24
arrived at top of hill at 3:30

Sonja [21]3 years ago

3 0

12 oclock to 6 oclock is 6 hours

d = distance
r = rate of speed

time to walk uphill : t1 = d/3
time to walk downhill: t2 = d/6
total distance = 2d ( up hill and down ill)
2d = (d/3 + d/6)r
2/(1/3 +1/6) = r
r = 4
average speed up and down the hill was 4 mph

total distance = (2/3 x 6 x 4) + (1/3 x 6 x 4)
total distance = 16 + 8 = 24 miles total

Average speed to top of hill = 2/(1/4 +1/3) = 3.43 mph

to top of hill = 24/2 = 12 miles

12 / 3.43 = 3.5 hours
12 oclock + 3.5 hours = 3:30

total miles was 24
arrived at top of hill at 3:30

You might be interested in

on the first day of every month Katie gets 450 phones she repairs 10 phones every day how many phones does she have left to repa

MaRussiya [10]

Answer:

340 phones

Step-by-step explanation:

Make an expression

She repairs 10 phones each day, d.

Each day she repairs 10 phones, it is 10 phones off the total number (450) she has to repair, so we can subtract 10d from 450.

450-10d

We want to find how many she has left after 11 days, so we can substitute 11 in for d

450-10(11)

Multiply

450-110

340

She will still have 340 phones left to repair

5 0

3 years ago

A group of physical education majors was discussing the heights of female runners and whether female runners tended to be tall,

Anna11 [10]

Answer:

The 90% confidence interval for the mean µ of the population of female runners.

( 65.0328 , 66.5672)

Step-by-step explanation:

Given A sample of 12 runners showed a sample mean height of 65.80 inches and a sample standard deviation of 1.95 inches.

Given sample size is n = 12 <30 so small sample

Given sample mean (x⁻) = 65.80 inches

sample standard deviation (S) = 1.95 inches.

Assume the population is approximately normal.

The 90% confidence interval for the mean µ of the population of female runners.

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )$

substitute all above interval

$(65.80 - t_{\alpha } \frac{1.95}{\sqrt{12} } ,65.80 +t_{\alpha } \frac{1.95}{\sqrt{12} } )$

The degrees of freedom γ=n-1 = 12-1=11

From t- table = 1.363 at 90 % 0r 0.10 level of significance

$(65.80 -1.363 \frac{1.95}{\sqrt{12} } ,65.80 +1.363 \frac{1.95}{\sqrt{12} } )$

on calculation , we get

(65.80 -0.7672 ,65.80 + 0.7672)

( 65.0328 , 66.5672)

8 0

3 years ago

Find 3 consecutive even integers if their sum decreased by 10 equals -22

Orlov [11]

Three consecutive even integers can be represented by x, (x+2), and (x+4).

(x + (x + 2) + (x + 4)) - 10 = -22 Get rid of the parentheses
x + x + 2 + x + 4 - 10 = -22 Combine like terms (x + x + x) and (2 + 4 - 10)
3x - 4 = -22 Add 4 to both sides
3x = -18 Divide
x = -6
Check you work.

x = -6
x + 2 = -4
x + 4 = -2

(-6 - 4 - 2) - 10 = 22
-12 - 10 = -22

So, your three integers are -6, -4, and -2 .

6 0

3 years ago

Please help me I am horrid at math!

Serga [27]

Answer:23 inches

<h2>Step-by-step explanation:all sides equal 23 when added together</h2>

7 0

3 years ago

Find the distance, to the nearest tenth, between (6, 5) and (-3, -4).

labwork [276]

Answer:

12.70

Step-by-step explanation:

7 0

3 years ago