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2 years ago

-3.25, The greater number is . The number farther from 0 is

Mathematics

1 answer:

astra-53 [7]2 years ago

3 0

25 and the farthest is also 24

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5 7/8 + 11/4<br> show me how to solve

loris [4]

let's firstly convert the mixed fractions to improper fractions and then add, bearing in mind that the LCD from 8 and 4 is simply 8.

$\bf \stackrel{mixed}{5\frac{7}{8}}\implies \cfrac{5\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{47}{8}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{47}{8}+\cfrac{11}{4}\implies \stackrel{\textit{using the LCD of 8}}{\cfrac{(1)47~~-~~(2)11}{8}}\implies \cfrac{47-22}{8}\implies \cfrac{25}{8}\implies 3\frac{1}{8}$

7 0

2 years ago

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In an airplane 35% of the passengers are children they’re 140 children on the airplane how many passengers are on the airplane

Angelina_Jolie [31]

Answer:

there are 49 passengers

Step-by-step explanation: 35%=0.35

0.35*140= 49

3 0

2 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

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

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

2 years ago

HELP ME PLZZZ!!!! URGENT!!!

finlep [7]

Answer:

Step-by-step explanation:

Given equation is,

x² + (p + 1)x = 5 - 2p

x² + (p + 1)x - (5 - 2p) = 0

x² + (p + 1)x + (2p - 5) = 0

Properties for the roots of a quadratic equation,

1). Quadratic equation will have two real roots, discriminant will be greater than zero. [(b² - 4ac) > 0]

2). If the equation has exactly one root, discriminant will be zero [(b² - 4ac) = 0]

3). If equation has imaginary roots, discriminant will be less than zero [(b² - 4ac) < 0].

Discriminant of the given equation = $(p+1)^2-4(1)(2p-5)$

For real roots,

$(p+1)^2-4(1)(2p-5)>0$

p² + 2p + 1 - 8p + 20 > 0

p² - 6p + 21 > 0

For all real values of 'p', given equation will be greater than zero.

5 0

2 years ago

(a + b - c )(a + b + c )

denis-greek [22]

Answer:

$a^2+b^2-c^2+2ab$

Step-by-step explanation:

$(a+b-c)(a+b+c)=a^2+ab+ac+ab+b^2+bc-ac-bc-c^2=a^2+b^2-c^2+2ab$

Hope this helps!

5 0

2 years ago

Read 2 more answers