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2 years ago

Which number is NOT in the solution set of the inequality x > 10?

Mathematics

1 answer:

Serjik [45]2 years ago

5 0

Answer:

Step-by-step explanation:

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Select all statements that show correct reasoning for finding 15 = 3

mario62 [17]

Answer:

. Multiply 15 by 1, then divide by 5

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2 years ago

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HURRY PLS!!!!! <br><br> What term can you add to 5/6 x - 4 to make it equivalent to 1/2 x - 4

77julia77 [94]

subtract 1/3x

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3 years ago

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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

3 years ago

A hemisphere-shaped bowl with radius 1 foot is filled full with chocolate. All of the chocolate is then evenly distributed betwe

Wewaii [24]

The radius of each of the smaller molds is r = 1/3 ft. Using the volume of the given hemisphere, the required radius is calculated.

<h3>How to calculate the volume of a hemisphere?</h3>

The volume of the hemisphere is calculated by using the formula,

= $\frac{2}{3}$ × π × r³ cubic units

where r is the radius of the hemisphere.

<h3>Calculation:</h3>

It is given that,

A hemisphere-shaped bowl with a radius r = 1 ft is filled with chocolate.

So, the volume of the bowl is

V = $\frac{2}{3}$ × π × (1)³

= $\frac{2}{3}$ × π cubic feets

The volume of each smaller hemisphere-shaped mold = $\frac{2}{3}$ × π × r³ cubic units

So, for 27 molds, the volume = 27 × $\frac{2}{3}$ × π × r³ cubic units

All of the chocolate is then evenly distributed between 27 congruent, smaller hemisphere-shaped molds.

Then,

$\frac{2}{3}$ × π = 27 × $\frac{2}{3}$ × π × r³

⇒ r³ = 1/27

⇒ r = 1/3 ft

Therefore, the required radius of each of the smaller molds is 1/3 foot.

Learn more about the volume of a hemisphere here:

brainly.com/question/15975126

#SPJ4

6 0

2 years ago

Evaluate the expression: (–2)2 + (–42) + (18 – 23).

marta [7]

Answer:

-51 (no options are matching)

Step-by-step explanation:

Given: (–2)2 + (–42) + (18 – 23).

we know, 2*2 = 4, and in integer multiplication if any negative integer is multiplied with positive integer, then the result will have a negative sign.

= -4 + (-42) + (18 – 23)

Here, we have open the brackets, and written the respective signs with the integers,

= - 4 - 42 - 5

now, we will perform the integer addition. Adding all the positives together & he negative numbers together

= -51

PS: It happens to be that none of the given options match with the correct answer. But, i have solved taking the expression, i hope it helps

6 0

3 years ago

Read 2 more answers