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Kristen can lift 133% of what Jane can lift. How many times more is that?

lubasha [3.4K]

The weight Kristen can lift is 1.3 times more than Jane's

<h3>How to determine the number of times?</h3>

The given paramater can be represented using the following equation

Kristen = 133% * Jane

Express the percentage as decimal

Kristen = 1.33 * Jane

Hence, the weight Kristen can lift is 1.3 times more than Jane's

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8 0

2 years ago

5x+25=2x+29 what’s the measures

AysviL [449]

the measures are x=4/3

4 0

4 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:

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

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

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

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

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

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

So

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

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

Write the length of the segments that connect each pair of points, in units.

MArishka [77]

The lengths of each of the segments connected by the given pairs of points are:

1. AB = 10 units

2. CD = 17 units

3. EF = 3 units

<h3>How to Find the Length of Segments Connected by Two Points?</h3>

To find the length of a segment connected by two coordinate points, the distance formula is applied, which is:

d = $\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$ .

1. Find the length of segment AB:

A(5,-3)

B(-3,3)

AB = √[(−3−5)² + (3−(−3))²]

AB = √[(−8)² + (6)²]

AB = √100

AB = 10 units

2. Find the length of segment CD:

C(-2, -7)

D(6, 8)

CD = √[(6−(−2))² + (8−(−7))²]

CD = √(64 + 225)

CD = 17 units

3. Find the length of segment EF:

E(5,6)

F(5,3)

EF = √[(5−5)² + (3−6)²[

EF = √(0 + 9)

EF = √9

EF = 3 units

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4 0

1 year ago

The area of a rectangle is 92 square centimeter. The length of the rectangle is 23 centimeters. Find the width of the rectangle.

bagirrra123 [75]

23 times 4= 91, so the width is 4

3 0

3 years ago

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