Solve for "x"4x-(2x 3)=7

What is the least common multiple of 12 and 10

gtnhenbr [62]

12 = 2*2*3

10 = 5*2

LCM = 5*3*2*2

= 15*4=60

the least common multiple is 60

4 0

4 years ago

Find the value of x<br> PIC IS BELOW

maks197457 [2]

: $\implies$ $\sf 3x = 27⁰$

: $\implies$ $\sf x = \dfrac{27}{3}$

: $\implies$ $\sf x = 9$

c.) x = 9⁰, angle measure is 27⁰

5 0

3 years ago

The figure below shows a tent with wires attached to help stabilize it. The length of each wire is 8 feet greater than the dista

Anestetic [448]

Answer:

The length of 2 wires are 13 feet and 12 feet [i don't know if the 3rd side is a wire, if so, the length is 5 feet]

Step-by-step explanation:

<em>The image of the tent is attached.</em>

<em />

As we can see, it creates a right triangle with "x+8" side being the hypotenuse of the triangle.

The pythagorean theorem tells us:

Hypotenuse^2 = Leg^2 + Leg^2

THe hypotenuse is "x+8", the two legs are "x" and "x+7" respectively. Let's solve for x:

$(x+8)^2=(x+7)^2+x^2$

We expand and solve for x:

$(x+8)^2=(x+7)^2+x^2\\x^2+16x+64=x^2+14x+49+x^2\\x^2+16x+64=2x^2+14x+49\\x^2-2x-15=0\\(x-5)(x+3)=0\\x=5,-3$

Length can't be negative, so x = 5.

Hence, the 3 sides of the triangle are:

x = 5

x + 8 = 13

x + 7 = 12

The length of 2 wires are 13 feet and 12 feet

8 0

3 years ago

how many different scores are possible if darts are thrown at the target and each dart lands within a ring. Target: 3,6,9

weeeeeb [17]

the only score is 18

7 0

4 years ago

Read 2 more answers

Candy. Someone hands you a box of a dozen chocolate-covered candies, telling you that half are vanilla creams and the other half

BaLLatris [955]

Answer:

a) P=0.091

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but it is still possible that there are 6 of each taste.

c) The probability of picking 4 vainilla in a row, if there are half of each taste, is P=0.030.

This is a very improbable case, so if this happens we would have reasons to think that there are more than half vainilla candies in the box.

Step-by-step explanation:

We can model this problem with the variable x: number of picked vainilla in a row, following a hypergeometric distribution:

$P(x=k)=\dfrac{\binom{K}{k}\cdot \binom{N-K}{n-k}}{\binom{N}{n}}$

being:

N is the population size (12 candies),

K is the number of success states in the population (6 vainilla candies),

n is the number of draws (3 in point a, 4 in point c),

k is the number of observed successes (3 in point a, 4 in point c),

a) We can calculate this as:

$P(x=3)=\dfrac{\binom{6}{3}\cdot \binom{12-6}{3-3}}{\binom{12}{3}}=\dfrac{\binom{6}{3}\cdot \binom{6}{0}}{\binom{12}{3}}=\dfrac{20\cdot 1}{220}=0.091$

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but is possible.

c) In the case k=4, we have:

$P(x=3)=\dfrac{\binom{6}{4}\cdot \binom{6}{0}}{\binom{12}{4}}=\dfrac{15\cdot 1}{495}=0.030$

This is a very improbable case, so we would have reasons to think that there are more than half vainilla candies in the box.

4 0

3 years ago