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vladimir2022 [97]
3 years ago
15

Find the missing fraction. ___ + 1/16 = 3/2 No links!

Mathematics
1 answer:
DochEvi [55]3 years ago
7 0

Answer:

23/16

Step-by-step explanation:

3*8/2*8=24/16-1/16=23/16

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Drag the tiles to list the sides of △MNO from shortest to longest.
sweet [91]

The smaller the angle subtended by a side, the smaller the length of the

side.

The correct responses are;

Question 1: The list of sides from shortest to longest are;

  • MO/Shortest MO/Medium and MO/Longest

a) <u>Friday</u>

b) <u>70 minutes</u>

c) <u>40%</u>

d) Yes<u>,</u> <u>the sum of the </u><u>mean</u><u> number of </u><u>minutes spent</u><u> on </u><u>aerobic</u><u> training and the mean number of minutes spent on </u><u>strength</u><u> training is equal to the mean </u><u>total</u><u> number of minutes spent </u><u>training.</u>

From the given diagram, we have, the measure of the third angle, ∠O, is

found as follows;

∠O = 180° - 54° - 61° = 65°

Therefore, ∠O = The largest angle

We get;

The longest side is opposite the largest angle, which gives;

The shortest side is the side opposite ∠N (54°)= \frac{}{MO}

The next shortest side is the side opposite ∠M(61°) = \frac{}{NO}

The longest side is the side opposite ∠O(65°) = \frac{}{MN}

a) The time spent training on Tuesday = 60 + 10 = 70 minutes

The time spent training on Thursday = 50 + 30 = 80 minutes

The time spent training on Friday = 45 + 40 = 85 minutes

Therefore, the day the athlete spent the longest total amount of time training is on <u>Friday</u>

b) The time spent training on Monday = 10 + 20 = 30 minutes

The time spent training on Wednesday = 20 + 15 = 35 minutes

Therefore, we get;

30, 35, 70, 80, and 85

The median total number of minutes the athlete spent training each day = <u>70 minutes</u>

<u />

c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120

The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300

The  percentage spent on strength training = \frac{120}{300} × 100 = \frac{40}%

d) The mean number of minutes spent on strength training is found as follows;

Mean_{strength} =\frac{120}{5} =24

The mean number of minutes spent on aerobic training is found as follows;

Mean_{aerobic} =\frac{10+60+20+50+40}{5} =36

Mean_{strength} +Mean_{aerobic} =24+36=60

The mean total number of minutes spent training, Mean_{total} = \frac{300}{5} = 60

Therefore;

  • Mean_{strength}+Mean_{aerobic} = Mean_{total} \\

Learn more here:

brainly.com/question/2962546

4 0
2 years ago
The volume of a packaging box is 5 - x cubic feet.The width of the box is x feet and the length is x - 2 feet. The height is giv
netineya [11]
A rational expression is undefined when the denominator is zero.
V=\dfrac{5-x}{x(x-2)}
is undefined when x=0 or x=2.
4 0
3 years ago
Read 2 more answers
If susan, janice, and melvin take 2.0 hours to walk to their school at a rate of 1.0 m/s, how far is their school from their hou
RideAnS [48]

Answer:

7.2 km

Step-by-step explanation:

Given that:

Hours taken = 2 hours

Walk rate = 1 m/s

Distance = Speed * time

Time taken in seconds = 2 * 60 * 60 = 7200 s

Hence,

Distance = 1 m/s * 7200s

Distance = 7200 meters

Distance in km = 7200 / 1000 = 7.2 km

3 0
2 years ago
Please help! Will mark brainliest!
Anestetic [448]

Each spinner has 5 numbers, the total combinations would be 5 x 5 = 25 combinations,

The answer is 25

6 0
3 years ago
A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate
Katyanochek1 [597]

Answer:

Therefore the rate change of height is  \frac{1}{\pi} m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

\frac{dV}{dt}=4 \ m^3/s

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is V=\pi r^2 h

                                                 =( \pi \times 2^2\times h)\ m^3

\therefore V= 4\pi h

Differentiating with respect to t.

\frac{dV}{dt}=4\pi \frac{dh}{dt}

Putting the value \frac{dV}{dt}

\Rightarrow 4\pi \frac{dh}{dt}=4

\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}

\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}

The rate change of height does not depend on the height.

Therefore the rate change of height is  \frac{1}{\pi} m/s.

4 0
3 years ago
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