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3 years ago

Please help I have a test tomorrow and I really need to study the pic is on top

Mathematics

2 answers:

In-s [12.5K]3 years ago

7 0

A. Who has the least amount saved? I think it's the answer

sp2606 [1]3 years ago

3 0

A: Anna has the least amount saved.

b:2,100 is the total amount of every bodies savings put together.

Since you already have 1,575 as info you can use that to divide by 15 because that is her ratio. If you divide it you get 105. You can use that for your other problems.

Anna has 210 dollars in savings. Beth has 315 dollars in savings. And Cynthia has 1,575 dollars in savings.

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Solving two step equations w/-3+14=5

natali 33 [55]

w/–3 + 14 = 5

⇛ w/–3 = 5 – 14

⇛ w/–3 = – 11

⇛ w = –11 × (–3)

⇛ w = 33

7 0

2 years ago

How is the graph of y=4(2)^x+3 translated from the graph of y=4(2)^x

Pie

Answer:

Vertical movement: Move up 3 units

Horizontal movement: Move left 3 units

Step-by-step explanation:

If your parent equation is $f(x) = 4(2)^{x}$ and your child equation is $f(x) = 4(2)^{x} + 3$ , then it has vertically moved up 3 units.

If your parent equation is $f(x) = 4(2)^{x}$ and your child equation is $f(x) = 4(2)^{x+3}$ , then it has moved horizontally left 3 units.

7 0

3 years ago

Read 2 more answers

Scott burned 20 calories per minute jogging for x minutes and 5 calories per minute for biking for y minutes. He spent a total o

stiv31 [10]

Answer:

x = 112

Step-by-step explanation:

We can create two equations 1. one that calculates calories burned and 2.another that calculates minutes:

20*x + 5*y = 1200

x + y = 120

Then we can solve for x

x = 120 - y

then we can plug that into the top equation:

20(120 - y) + 5y = 1200

divide by 5:

4(120 - y) + y = 240

distribute:

480 - 4y + y = 240

and then do all that combining like factors what not and simplify and solve:

480 - 3y = 240

-3y = -240

y = 8

NOW WE CAN JUST PLUG IN THE ANSWER FOR Y TO FIND X

8+x =120

x = 112

7 0

3 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 ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$