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Zielflug [23.3K]

3 years ago

8

Can you show a break down of the Distributive Property for 5x198

1 answer:

nikitadnepr [17]3 years ago

4 0

5x100+5x90+5x8 and there you go all you need to do is do this problem the right way and you will get the right answer

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Sami leave his house at 3:45 PM and walks to karlas house in 15 minutes he stays at karlas house for five minutes and then walks

chubhunter [2.5K]

Sami arrives at the library at 4:15. Why? Well is says she took 15 minutes to walk to karlas house so when she arrives it’s 4:00 PM. Than, she stays at karlas house for 5 minutes which puts her at 4:05 PM. After that is Sami took 10 minutes to get to the library so that puts Sami at 4:15 PM. Hope this helps!

6 0

3 years ago

On a coordinate plane, a red curved line with an upward arc, labeled g of x, crosses the x-axis at (negative 2, 0), and the y-ax

maxonik [38]

Answer:

f(0) = g(0)

Step-by-step explanations

if a function h(x) passes through [a,b] then it can be said that h(a)=b.

given that, g(x) passes through (-2,0) and (0,4)

so, g(-2)=0, g(0)=4.

f(x) passes through (0,4) and (2,0)

so, f(2)=0, f(0)=4.

therefore, f(0)=g(0)=4.

so, first option is correct.

7 0

4 years ago

Read 2 more answers

With solving and it’s better to edit the photo and send it

Daniel [21]

Step-by-step explanation:

please mark me as brainlest

5 0

3 years ago

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

5 0

3 years ago

Need help on this. Don’t understand a thing and I need to get this done ASAP. If you knew please help. It would be greatly appre

matrenka [14]

Answer:

the Lateral SA is 84

The total is 1092.

Step-by-step explanation:

4 0

3 years ago