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wolverine [178]
3 years ago
6

Simplify the ratio 2:10

Mathematics
1 answer:
Tamiku [17]3 years ago
5 0
The answer to this question is 1:5
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Kobotan [32]
I hope this helps you




volume of sphere =4/3.pi.r^3
6 0
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6.493 rounded to the hundredths
Nana76 [90]
The answer is 6. 49 you welcome :)
7 0
3 years ago
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A pair of fair dice is cast. Let Edenote the event that the number landing uppermost on the first die is a 1 and let Fdenote the
iris [78.8K]

Answer:

They are not independent

Step-by-step explanation:

Given

E = Occurrence of 1 on first die

F = Sum of the uppermost occurrence in both die is 5

Required

Are E and F independent

First, we need to list the sample space of a roll of a die

Event\ 1 = \{1,2,3,4,5,6\}

Next, we list out the sample space of F

Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\

6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}

In (1): the sample space of E is:

E = \{1\}

So:

P(E) = \frac{n(E)}{n(Event\ 1)}

P(E) = \frac{1}{6}

In (2): the sample space of F is:

F = \{5,5,5,5\}

So:

P(F) = \frac{n(F)}{n(Event\ 2)}

P(F) =\frac{4}{36}

P(F) =\frac{1}{9}

For E and F to be independent:

P(E\ and\ F) = P(E) * P(F)

Substitute values for P(E) and P(F)

This gives:

P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}

P(E\ and\ F) = \frac{1}{54}

However, the actual value of P(E and F) is 0.

This is so because E = \{1\} and F = \{5,5,5,5\} have 0 common elements:

So:

P(E\ and\ F) = 0

Compare P(E\ and\ F) = \frac{1}{54} and P(E\ and\ F) = 0.

These values are not equal.

Hence: the two events are not independent

6 0
3 years ago
Part A. Jose painted a square shaped mural. The length of each side of the canvas is 1 1/3 yards. What is the area of the mural
Naily [24]
A. Its a square. This means each side times the side would give us the area (s^2)
4/3^2 or 1.3333 x 1.33333 = 16 / 9 area.

B. Really just pick any numbers you want.
16 yards long and 1/9 wide.
As long as they multiply to 16/9
4 0
3 years ago
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The value V of a Porsche 718 Cayman that is t years old can be modeled by V(t) = 420,000(0.965)t (a) What would be worth the car
nordsb [41]

Answer:

<u>A. V (2) = $ 391,114.50</u>

<u>B. t = 7.2 years (rounding to the next tenth) or approximately 7 years, 2 months and 12 days  </u>

Step-by-step explanation:

Given formula:

V(t) = 420,000 * (0.965)^t

A. What would be worth the car′s worth in 2 years?

V(t) = 420,000 * (0.965)^t

We replace t by 2, as follows:

V(2) = 420,000 * (0.965)²

V(2) = 420,000 * 0.931225

<u>V (2) = $ 391,114.50</u>

B. In how many years will the car be worth $325,000?

V(t) = 420,000 * (0.965)^t

We replace V(t) by 325,000, as follows:

325,000 = 420,000 * (0.965)^t

325,000/420,000 = (0.965)^t

0.77381 =  (0.965)^t

t = log 0.965(0.7738)  

t = log 0.7738/log 0.965  

<u>t = 7.2 years (rounding to the next tenth) or approximately 7 years, 2 months and 12 days  </u>

0.2 years = 0.2 * 12 = 2.4 months

0.4 months = 0.4 * 30 = 12 days

7 0
3 years ago
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