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3 years ago

How to round 1.61 to the nearest ten?

Mathematics

1 answer:

MakcuM [25]3 years ago

7 0

Answer:

Step-by-step explanation:

tens are 0, 10, 20, 30, etc

it's between 0 and 10 but closer to 0

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4:15 to 5 :00 is how many minutes

Inga [223]

45 Minutes. Time is like subtraction when you want to figure it out :) Hope this helps~!

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3 years ago

What is the eighth term in the pattern 2,10, 18,26,

Oksana_A [137]

The 8th character would be 58

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3 years ago

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution 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.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

3 years ago

A population of 200 animals has a growth rate of 1.03% each year. at this growth rate, the function f(x)=200(1.03)^x gives the p

Y_Kistochka [10]

For this case, the first thing we should do is use the following equation:
f (x) = 200 (1.03) ^ x
We substitute f (x) = 300:
300 = 200 (1.03) ^ x
We clear the value of x:
log1.03 (300/200) = log1.03 ((1.03) ^ x)
x = log1.03 (300/200)
x = 13.72
Answer:
the population will first reach 300 in about 13.72 years.

6 0

3 years ago

Construct a truth table for the statement ~( p v ~q) <br> pq Truth Table

Likurg_2 [28]

Hello

T for True

F for False

$\left|\begin{array}{c|c|c|c|c}p&q&\~q&pv\~q&\~(pv\~q)\\---&---&---&---&---\\T&T&F&T&F\\T&F&T&T&F\\F&T&F&F&T\\F&F&T&T&F\\\end{array}\right|$

thanks

4 0

3 years ago