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What multiplication fact can you double to find 4x7

AnnZ [28]

You can do subtract half of 4 which is 2 and multiply by 2 by 7 so it will be 2x14 which is the same thing as 7x4

5 0

2 years ago

Read 2 more answers

4) y = Justify: 63° у

kenny6666 [7]

Answer:

y = 63° because they are alternate angles.

3 0

2 years ago

The values of 60, 62, and 84 were common to both samples. The three values are identified as outliers with respect to the age-gr

postnew [5]

Answer:

Only 60 is an identified outlier

Step-by-step explanation:

Given

Age group: 40 to 50

$Min = 60$ $Q_1 = 70$ $Median= 73$

$Q_3 = 76$ $Max = 85$

Conditions for outlier

$Value > Q_3 + 1.5 * IQR$

$Value < Q_1 - 1.5 * IQR$

See attachment for complete question

Required

Which of 60, 62 and 84 is an outlier of age grout 40 to 50 years

First, calculate the IQR of group 40 to 50.

This is calculated as:

$IQR = Q_3 - Q_1$

Where:

$Q_3 = 76$ and $Q_1 = 70$

So:

$IQR = 76 - 70$

$IQR = 6$

Next, is to test the outlier conditions on each value (i.e. 60, 62 and 84)

Testing 60

Condition 1

$Value > Q_3 + 1.5 * IQR$

$60 > 76 + 1.5 * 6$

$60 > 76 + 9$

$60 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$60 < 70 - 1.5 * 6$

$60 < 70 - 9$

$60 < 61$ --- True

Because one of the conditions is true, then 60 is an outlier of group 40 - 50 years

Testing 62

Condition 1

$Value > Q_3 + 1.5 * IQR$

$62> 76 + 1.5 * 6$

$62 > 76 + 9$

$62 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$62 < 70 - 1.5 * 6$

$62 < 70 - 9$

$62 < 61$ --- False

Because both conditions are false, then 62 is not an outlier of group 40 - 50 years

Testing 84

Condition 1

$Value > Q_3 + 1.5 * IQR$

$84> 76 + 1.5 * 6$

$84 > 76 + 9$

$84 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$84 < 70 - 1.5 * 6$

$84 < 70 - 9$

$84 < 61$ --- False

Because both conditions are false, then 84 is not an outlier of group 40 - 50 years

5 0

2 years ago

20 POINTS WILL GIVE BRAINLIEST plz help with Question below

Anuta_ua [19.1K]

Answer:

Graph of Option D represents $y=\sqrt{x}$

Step-by-step explanation:

we are given our function as

$y=\sqrt{x}$

squaring on both sides we get

$y^2=x$

It represent a parabola opening towards the positive side of x axis. Hence it gives us some preliminary idea about the graph of the function we are given .

However our original function is $y=\sqrt{x}$

Domain of $y=\sqrt{x}$ is all positive values of x

And square root of positive values will always result in positive values. Hence y can not be negative as the range is All positive values of y

Hence we erase the graph of $y^2=x$ below x axis to obtain the graph of

$y=\sqrt{x}$

3 0

2 years ago

Bacteria of species A and species B are kept in a single environment, where they are fed two nutrients. Each day the environment

DiKsa [7]

Answer:

We require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

Step-by-step explanation:

Let n₁ be the population of A required and n₂ be the population of B required.

Now we require 2 units of the first nutrient for species A and one unit of the first nutrient for species B. The total nutrients required by species A is 2n₁ and that by species B is 1n₂ = n₂. So, the total nutrients required by both species A and B is 2n₁ + n₂. Since this equals the quantity of the first nutrient which is 10,560, then 2n₁ + n₂ = 10,560 (1)

Now we require 5 units of the second nutrient for species A and 6 units of the second nutrient for species B. The total nutrients required by species A is 5n₁ and that by species B is 6n₂. So, the total nutrients required by both species A and B is 5n₁ + 6n₂. Since this equals the quantity of the first nutrient which is 31,510, then 5n₁ + 6n₂ = 31,510 (2).

So, we have two simultaneous equations which we would solve to find the populations of A and B which satisfy both equations.

2n₁ + n₂ = 10,560 (1)

5n₁ + 6n₂ = 31,510 (2)

From (1) n₂ = 10,560 - 2n₁ (3)

Substituting equation (3) into (2), we have

5n₁ + 6(10,560 - 2n₁) = 31,510

expanding the brackets, we have

5n₁ + 63,360 - 12n₁ = 31,510

collecting like terms, we have

5n₁ - 12n₁ = 31,510 - 63,360

simplifying, we have

- 7n₁ = -31,850

dividing both sides by -7, we have

n₁ = -31,850/-7

n₁ = 4,550

Substituting n₁ = 4,550 into (3), we have

n₂ = 10,560 - 2(4,550)

n₂ = 10,560 - 9,100

n₂ = 1,460

So, we require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

3 0

1 year ago