All categories

3 years ago

Please help asap 30 pts

Mathematics

2 answers:

aliya0001 [1]3 years ago

8 0

$5y+1.8=4y-3.2\qquad|\text{subtract 1.8 from both sides}\\\\5y=4y-5\qquad|\text{subtract 4y from both sides}\\\\\boxed{y=-5}\to\boxed{a.}$

Semmy [17]3 years ago

4 0

Step 1. Subtract 4y from both sides

5y + 18 - 4y = -3.2

Step 2. Simplify 5y + 1.8 -4y to y + 1.8

y + 1.8 = -3.2

Step 3. Subtract 1.8 from both sides

y = -3.2 - 1.8

Step 4. Simplify -3.2 - 1.8 to -5

y = -5

Your answer is A.

You might be interested in

Which graph below shows a line that contains ordered pairs that are solutions of the equation shown?

Vinvika [58]

Given:

The equation is:

$y=-x+4$

To find:

The graph of the line that contains ordered pairs that are solutions of the given equation.

Solution:

We need to find the graph of the given equation.

We have,

$y=-x+4$

At $x=0$ , we get

$y=-0+4$

$y=4$

So, the x-intercept of the given equation is at point (0,4).

At $y=0$ , we get

$0=-x+4$

$x=4$

So, the y-intercept of the given equation is at point (4,0).

From the given graphs it is clear that the line in option C has x-intercept at point (0,4) and y-intercept at point (4,0).

Therefore, the correct option is C.

7 0

3 years ago

Which point lies on the graph of the equation 10y = 3x - 11?

GaryK [48]

Answer:

A. (2,-0.5)

Step-by-step explanation:

A pair of coordinates is always (x,y) so

If you insert the numbers into the equation you get

10(-0.5) = 3(2) - 11

-5 = 6 - 11

-5 = -5

Therefore (2,-0.5) is the point that likes on the graph

5 0

3 years ago

70% of what number is 56?

sashaice [31]

It’s 80 !!!!!!!!!!! Your welcome love

4 0

3 years ago

Read 2 more answers

In a widget factory, machines A, B, and C manufacture, respectively, 20, 30, and 50 percent of the total. Of their output 6, 3,

steposvetlana [31]

Answer:

38.71% probability it was manufactured by machine A.

29.03% probability it was manufactured by machine B.

32.26% probability it was manufactured by machine C.

Step-by-step explanation:

We have these following probabilities:

A 20% probability that the chip was fabricated by machine A.

A 30% probability that the chip was fabricated by machine B.

A 50% probability that the chip was fabricated by machine C.

A 6% probability that a chip fabricated by machine A was defective.

A 3% probability that a chip fabricated by machine B was defective.

A 2% probability that a chip fabricated by machine C was defective.

The question can be formulated as:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

What are the probabilities that it was manufactured by machines A, B, and C?

Machine A

What is the probability that the widget was manufactures by machine A, given that it is defective?

P(B) is the probability that it was manufactures by machine A. So $P(B) = 0.20$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine A. So $P(A/B) = 0.06$

P(A) is the probability that a widget is defective. This is the sum of 6% of 20%, 3% of 30% and 2% of 50%. So

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.2*0.06}{0.031} = 0.3871$

38.71% probability it was manufactured by machine A.

Machine B

What is the probability that the widget was manufactures by machine B, given that it is defective?

P(B) is the probability that it was manufactures by machine B. So $P(B) = 0.30$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine B. So $P(A/B) = 0.03$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.03}{0.031} = 0.2903$

29.03% probability it was manufactured by machine B.

Machine C

What is the probability that the widget was manufactures by machine C, given that it is defective?

P(B) is the probability that it was manufactures by machine C. So $P(B) = 0.50$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine C. So $P(A/B) = 0.02$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.02}{0.031} = 0.3226$

32.26% probability it was manufactured by machine C.

6 0

3 years ago

A high school baseball player has a 0.215 batting average. In one game, he gets 7 at bats. What is the probability he will get a

masya89 [10]

Answer:

0.651%

Step-by-step explanation:

using Binomial expansion: nCr*p^n*(1-p)^(n-r)

probability he will get at least 5 hits in the game = probability he will get 5 hits in the game + probability he will get 6 hits in the game + probability he will get 7 hits in the game = 7C5*0.215^5*(1-0.215)^2 + 7C6*0.215^6*(1-0.215)^1 +7C7*0.215^7*(1-0.215)^0 = 21* 0.0004594*0.616225 + 7*0.00009877*0.785 + 1*0.000021235 = 0.00651 = 0.651%

7 0

3 years ago