All categories

3 years ago

Convert the number 243FIVE to base ten

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Step-by-step explanation:

ok that is the answer bro

You might be interested in

The ratio of apples to bananas in a bowl of fruit is 2:5 if there are 49 total pieces of fruit in the bowl how many are apples

Ghella [55]

14 apples because altogether the ratio adds up to 7 and 49/7=7 so we could rewrite the ratio as 2x:5x so x=7. there for 2x ,the number of apples is 14

7 0

3 years ago

40% of what number is 28

Leto [7]

70
0.4x=28
x=28/0.4
x=70

7 0

3 years ago

Read 2 more answers

Translate the figure 3 units right and 7 unit down.Record the coordinates of the image.

Firlakuza [10]

Just move over each of those points to theme right 3 and down 7 then connect the dots again

7 0

3 years ago

Read 2 more answers

According to a height and weight chart, Bruce's ideal wrestling weight is 168 pounds. Bruce

jek_recluse [69]

Super simple

So he wants to be three pounds away from his ideal weight (168)

Which means he can either be 3 pounds under or 3 pounds over

So his min is 165, his max is 171

5 0

3 years ago

For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let X = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in n missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

6 0

3 years ago