All categories

2 years ago

Help? pleaseee

Mathematics

2 answers:

Lapatulllka [165]2 years ago

8 0

Answer:

29.03 kg

Step-by-step explanation:

G00gle is your friend lol. 64lb = 29.0299117 kg

Rounded: 29.03kg

Sholpan [36]2 years ago

7 0

Answer: 29.03

Step-by-step explanation:

You might be interested in

if one more freaking person tell me that 120 / 37 is 3 with a decimal I'm going to delete this app because it is just three

lawyer [7]

But 37 is not a factor of 3, in fact it is a prime number so anything that is divisible by it (other than 1) will be a decimal. So, my friend here you have another person on this app telling you that it is not 3 and is in fact 3.2432 (;;;

4 0

3 years ago

Read 2 more answers

Three different whole numbers add up to 31.

lubasha [3.4K]

Let's find the least possibilities:

First number: 6
Second number: 8
Third number: 10

6 + 8 + 10 = 24

Now we can see that we are still missing 31 - 24 = 7.

7 can be gained by adding 3 and 4, so:

First number: 6 + 3 = 9
Second number: 8 + 4 = 12
Third number: 10

9 + 12 + 10 = 31

5 0

2 years ago

Tell whether the ordered pair is a solution of the equation<br> y=5x; (15,-3)

n200080 [17]

Answer:

This ordered pair is NOT a solution of the equation.

Step-by-step explanation:

First, it is important to note that ordered pairs consist of (x, y) so in this case

x = 15 and y = -3. And that any variable directly next to a number calls for multiplication. So, plugging in the variables the equation would look like

"-3 = 5 x 15"

To see if 5 x 15 = -3 we will multiply them. 5 x 15 actually equals 75 not -3. So we know this ordered pair is NOT the solution for the problem.

3 0

2 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

3 years ago

What is the anwser to 2y-4x=8

True [87]

The answer is y=2x+8 because your original is y=my+b

5 0

3 years ago

Read 2 more answers