Please help and i will give you the brainleist, please write the answer i notebook and send.

A donkey eats 3/9 of a hay bale a week. How many hay bales does he eat in 21 days?

masya89 [10]

Answer:

1

Step-by-step explanation:

since a week is 1/3 of 21 days. ALl we need to do is 3/9 x 3 which =1

3 0

3 years ago

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Plss help me with my question I'll give you more points

umka21 [38]

If you place the numbers that you have into the chart below and just move the decimal to where’s its asking you to, you’ll get the answers necessary

7 0

3 years ago

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The area of the cross section of a sphere at the largest point is 100π square miles. what is the total surface area of this sphe

Firlakuza [10]

The cross-section of a sphere is a circle . The area of a circle can be calculated by the formula A= $\pi\r^{2}$ Where r is the radius of the circle.

Given : $\pir^{2} = 100\pi$

Solving for r we have:

$r^{2} = \frac{100\pi }{\pi}= 100$

Or r= 10.

The area of a sphere is calculated by the formula :

$SA = 4\pi R^{2}$

Substituting r value we have:

SA= $4\pi\ 10 ^{2} = 400\pi$

The total surface area of this sphere is 400π .

4 0

3 years ago

53% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

faltersainse [42]

Answer:

a

$P(X = 5) = 0.2423$

b

$P( X \ge 6 )= 0.4516$

c

$P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.53

The sample size is n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as

$P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=> $P(X = 5) = 252 * 0.04182 * 0.023$

=> $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

$P( X \ge 6 )= P( X = 6) + P( X = 7) + P( X = 8) + P( X = 9) + P( X = 10)$

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

$P( X < 4 )= P( X = 3) + P( X = 2) + P( X = 1) + P( X = 0)$

=> $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$