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3 years ago

7

bianca is building a sandbox that will hold 4 cubic yards of sand. she wants it to be 18 inches deep and 2 yards wide. what will

be the length

1 answer:

tamaranim1 [39]3 years ago

5 0

Length should be 2/5yd:

18inch*1yd/36inch=.5yd=> depth

volume= 2(length*width+lenght*depth+depth*width)

let l to be the length.

volume= 2(l*2+2(.5)+.5*2)

volume=5l+2

4=5l+2

2=5l

l=2/5

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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

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2 years ago

Anthony puchased a coffee mug on sale for 20% off.

andrew-mc [135]

Answer:

6 dollars

Step-by-step explanation:

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Jonas has 40 marbles. He gave 1/5 of the marbles to marvin and 3/8 of the marbles to leroy. How many marbles did each boy get?

MaRussiya [10]

1/5 x 40

= 8 marbles

3/8 x 40

= 15 marbles

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2 years ago

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Sari drives 55 miles per hour. Which equation shows the proportional relationship between hours (h) and miles (m)?

Finger [1]

Answer:

$m=55h$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

m ----> the distance in miles (y-coordinate)

h ----> the time in hours (x-coordinate)

In this problem the constant of proportionality k is given and is equal to

$k=55\ \frac{miles}{hour}$ ----> the speed

substitute

$m=55h$

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3 years ago

To find the interquartile range you need to ____the upper and lower quartile values.

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Answer:

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2 years ago

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