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3 years ago

Solve for x. x 4 = 3 8

Mathematics

2 answers:

geniusboy [140]3 years ago

6 0

Answer:

x=6

Step-by-step explanation:

( 8 ) ⋅ ( 3 ) = ( x ) ⋅ ( 4 )

Cerrena [4.2K]3 years ago

3 0

Answer:

The answer is 9.5

Step-by-step explanation:

The first thing you do is you divide 38 by 4.

When you do that you get your answer which is 9.5

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Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

3 years ago

Help to solve this word problem?

goldfiish [28.3K]

Let's assume

length =l

width =w

we are given

the width of a rectangle is 7 feet less than its length

so, we get

$w=l-7$

now, we know that

area = length*width

$A=l*w$

now, we can plug w

$A=l*(l-7)$

we have

Area=170

$170=l*(l-7)$

$l^2-7l-170=0$

now, we can factor it

$(l-17)(l+10)=0$

now, we can solve for l

$(l-17)=0$

$length=17 feet$

now, we can find width

$w=17-7$

$width=10 feet$

so, dimensions are

$length=17 feet$

$width=10 feet$ ...............Answer

5 0

3 years ago

Please answer this.. please

Mice21 [21]

The second one and the last one

7 0

4 years ago

11. PARKING The rates at a short-term parking garage are $5.00 for 2 hours or less,

Vikentia [17]

I dont know ok ok ok ok ok ok ok ok ok ok

5 0

3 years ago

please help I have no idea on how to do this please help I will give a brainlyist to the first person to answer

Julli [10]

Cutting to the chase, the answer is more than 1 kilometer, up to 1,715 meters. If the speed of sound is 343 meters per second then 343 * 5 = 1,715 meters every 5 seconds

8 0

3 years ago