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Y_Kistochka [10]
2 years ago
7

In Centerville, 45% of the population is female, and 60% of the population commutes to work daily. Of the total Centerville popu

lation, 21% are females who commute to work daily. What percentage of the total Centerville population are males who do NOT commute to work daily?
Mathematics
2 answers:
Rzqust [24]2 years ago
8 0

Answer:

16%

Step-by-step explanation:

100%-45%=55%

60%-21%=39%

55%-39%=16%

the total Centerville population are males who do NOT commute to work daily is 16%.

tamaranim1 [39]2 years ago
3 0

Answer:

16%

Step-by-step explanation:

100%-45%-(60%-21%)=16%

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The sum of 2 numbers is 44 the smaller number is 16 less than the larger number what are the numbers
Anon25 [30]
X=27 and y=11
if you let each number equal x and y we know the sum is 38 so x+y=38
and the smaller is 16 less than the larger
. so if we say the smaller number is 7 then x-16=y
these can be solved as a simultaneous equation. x+(x-16)=38
2x-16=38
2x=54
x=27
(27)-16=y
y=11
so the answer is x=27 y=11
4 0
2 years ago
Twenty four over the quantity of 3 x minus two. Find f(−2).
TEA [102]
Answer:

f(-2) = -3

Explanation:

Not sure if those last couple numbers are answer choices, but I'm going to infer that they might be.

Twenty four over the quantity of 3 x minus two in standard form is:

f(x) = <span>24<span>3x−2</span></span>

Since the number in the ( ) = x, plug in for x using f(-2)

<span>24<span>3∗<span>(−2)</span>−2</span></span>

P E {M D} {A S} for where to solve first

Multiply the 3 and (-2):

<span>24<span>−6−2</span></span>

Add -6 and -2:

<span>24<span>−8</span></span>

Divide the rest:

f(-2) = -3

3 0
3 years ago
Ribbon candy costs $2.50 per oot. how many can you buy if you have $11.25?
ryzh [129]

Answer: 4 times

Step-by-step explanation:

2.50 x 4 = 10

I tried different numbers

5 0
3 years ago
Help plzz division math prob
leva [86]

g(x) = x - 3 = 0

g(x) = x = 3

f(x) = 2x^3 + x - 4

f(3) = 2(3)^3 + 3 - 4

f(3) = 2(27) - 1

f(3) = 54 - 1

f(3) = 53

The remainder when f(x) is divided by x - 3 is <u>53</u>.

3 0
2 years ago
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
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