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Dvinal [7]
3 years ago
12

Drag the phrases into the box to match each expression.

Mathematics
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

3 /4 x+ 1 /2 ​ one-half more than three-fourths of a number

3/ 4 − 1 /2 x  three-fourths minus one-half of a number

3/ 4 −(x+ 1/ 2 )  three-fourths minus the sum of a number and one-half



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Help on number 6 i don't understand how to do it and i got corrections so i need help to do them.
Mademuasel [1]
The answer #6 is j. 100. He uses the treadmill for 40% so that means 60% of his exercise is not on the treadmill. 

40%= 40 mins 
60%= 60 mins
 total: 100%= 100 mins.
Hope this helps!
8 0
3 years ago
Read 2 more answers
The riverside geyser in Yellowstone erupts about every 6.25 h. When the geyser erupts, the water has an initial upward velocity
Aleonysh [2.5K]

Answer:

Correct answer:  Hmax = 22.11 m

Step-by-step explanation:

The subject of this problem is a vertical shot upwards, to which the following formula applies:

V₀² - V² = 2 · g · H   where is:

V₀ - initial velocity

V - final velocity

g = 10 m/s²

H - final height

Given:

V₀ = 69 ft/s = 21.03 m/s

V = 0 m/s

Hmax = ?

V₀² = 2 · g · Hmax  ⇒ Hmax = V₀² / 2 · g

Hmax = 21.03² / 2 · 10 = 442.26 / 20 = 22.11 m

Hmax = 22.11 m

God is with you!!!

6 0
2 years ago
Pls help fast:<br>How many ways can 3 songs be chosen from 12 to create a 3 song ordered play list?​
leva [86]

Answer:

1320

Step-by-step explanation:

Total number of songs = 12

Number of songs required to arrange in a play list = 3

It means we have to arrange 3 songs from 12 in the play lines. So, required number of ways is

Ways=^{12}P_{3}

Ways=\dfrac{12!}{(12-3)!}         [\because ^nP_r=\dfrac{n!}{(n-r)!}]

Ways=\dfrac{12\times 11\times 10\times 9!}{9!}

Ways=1320

Therefore, the required number of ways is 1320.

3 0
3 years ago
Which equation represents a line which is parallel to the line 6y - 5x = -6?
Pavel [41]

Answer:

5

Step-by-step explanation:

6y - 5x = - 6

6y = 5x - 6

If the num beside the x is not 5, then it's not a parallel line

5 0
2 years ago
Given the probability distributions shown to the​ right, complete the following parts.
Elan Coil [88]

Answer:

a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

Variance = Var(X) = 10.46 - 3.02² = 1.3396

Standard deviation = √(1.3396) = 1.157

For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

7 0
3 years ago
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