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4 years ago

I need help with this three problems (with procedure)

Mathematics

1 answer:

finlep [7]4 years ago

7 0

Answer:

13. $x^{3} y^3$ cubic units

14. 43 million; 79 million; 3.6 × $10^7$

15. 2.68 × $10^5$

Step-by-step explanation:

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HELP PLEASE!

Svet_ta [14]

Hello! I believe the answer to your question is C (hours). If t is the amount of time it takes, hours is a unit of time, therefore I assume that would be your answer. :)

7 0

4 years ago

Write the equation of the parabola in vertex form<br>

damaskus [11]

Answer:

f(x)= - 4 ( $x^{2}$ ) + 3

Step-by-step explanation:

1. start with f(x)=a( $x-p)^{2}$ +q

2. in that vertex is (p,q) noticing that use given vertex(0,3) and change the equation/function --> f(x)=a(x-0)+3 --> f(x)=a( $x)^{2}$ +3

3. Then use the give point and substitute to find "a" value --> - 1 = a( $1)^{2}$ +3

--> - 1 = a + 3 --> a= - 4

4. put the "a" value into the #2. function/equation --> f(x)= - 4( $x)^{2}$ +3

5 0

3 years ago

If a game consists of picking a number from a bag of beans numbered 2, 2, 3, 4, and 5 and flipping a coin, which probabilities a

Semenov [28]

The true statement is (d) none of the above

<h3>Missing information</h3>

I. P(3 and tail)

II. P(even and head)

III .P(odd and head)

<h3>How to determine the equivalent probabilities?</h3>

We start by calculating each probability

There is only one 3 in the 5 numbers.

So:

P(3) = 1/5

The probability of a tail in a coin is:

P(Coin) = 1/2

So, we have:

P(3 and tail) = 1/5 * 1/2

P(3 and tail) = 1/10

There are three even numbers in the 5 numbers.

So:

P(even) = 3/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(even and head) = 3/5 * 1/2

P(even and head) = 3/10

There are two odd numbers in the 5 numbers.

So:

P(odd) = 2/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(odd and head) = 2/5 * 1/2

P(even and head) = 1/5

By comparing the results, none of the probabilities are equal.

Hence, the true statement is (d) none of the above