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3 years ago

This recipe calls for 12 cinnamon rolls but I need 30 what number do I multiply

Mathematics

2 answers:

Paha777 [63]3 years ago

7 0

Answer:

12 x 2.5 = 30

Step-by-step explanation:

aev [14]3 years ago

3 0

Answer:

1.5.Since normal quantites make 12 rolls ,double will make 24 and half of the ingredients will make only 6 rolls..making the number 30..Hence the number 1.5.

So just multiply each quantity by 1.5

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Please help me out !! ( I will mark brainliest)

lisov135 [29]

Question 1: The y-intercept is where the line crosses the y-axis. The increments of the y-axis is by 20, and the y-intercept is at (0, 20).

The slope is the change in y over the change in x. Find two points:

(0, 20) and (2, 80)

Now:

(80 - 20)/(2 - 0) = 60/2 = 30

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

So:

y = 30x + 20

Question 2: Again, look for the y-intercept, which is pretty clear. It's (0, 60).

Now find two points:

(0, 60) and (2, 40)

Find the slope:

(40 - 60)/(2 - 0) = -20/2 = -10

So, the equation is:

y = -10x + 60

And, there you go!

3 0

3 years ago

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

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

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

3 years ago

What is the probability of flipping a coin and getting tails three times in a row

Artemon [7]

If im not mistaken its a 1/8 probability

7 0

3 years ago

Q = m-a/x (solve for m)

DaniilM [7]

Answer:

m=a/x+Q

Step-by-step explanation:

Q=m-a/x

To solve this we need to isolate m so we will bring -a/x to the other side of the equal sign so it will become positive i.e. ax now will be written as a/x+Q=m

in other wards m=a/x+Q

7 0

3 years ago

Answer fast for pts

sdas [7]

Answer:

Since she arrived at 8:30 its not accurate because her records started at the time she arrived so she didnt get the times of the students that arrived before her

Step-by-step explanation:

Hope this helps:))

4 0

2 years ago

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