What is MULTIPLIED to equal 80 but can be ADDED to make NEGATIVE 21

MakcuM [25]

X = measure of angle 1
y = measure of angle 2
z = measure of angle 3
w = measure of angle 4

Focus on the bottom triangle. The three angles add to 180 degrees
(angle 2) + (angle 3) + 116 = 180
y+z+116 = 180
y+z= 180-116
y+z= 64

Since we have the bottom triangle as isosceles, this means that y = z, so
y+z = 64
y+y = 64
2y = 64
y = 64/2
y = 32
making z = 32 as well

Similarly, angle 1 and angle 4 are 32 degrees because the 116 angle is opposite the top left-most angle, and congruent to this angle. In other words, the bottom triangle is a mirror image of the top triangle.

The figure is a rhombus because all four sides are the same length (as shown by the tickmarks)

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Answer:
This figure is a rhombus
All four angles (angle 1 through angle 4) are the same measure. They are each 32 degrees

6 0

3 years ago

What is the value of M?<br> Step by step explanation thanks!

Free_Kalibri [48]

M = 72

Your main goal in this equation is to isolate m or in other words leave m alone in one side of the equality.

All you need to do is multiply 4 in both sides for you to eliminate the denominator attached to m.

Saying this,

(M/4)(4) = (18)(4)

M = 72

4 0

3 years ago

According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se

belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers 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.

50% of adult smokers started smoking before 21 years old.

This means that $p = 0.5$

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that $n = 5$ .

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

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

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

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875$

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

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

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875$

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

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

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

4 0

2 years ago

Find the x-intercepts (solve for x) for the following AND PLEASE SHOW YOUR WORK!

alekssr [168]

Answer:

hope it may help but I wasnt able to find y cause no question

4 0

3 years ago

With a short time remaining in the day, a delivery driver has time to make deliveries at 5 locations among the 9 locations remai

Delvig [45]

Answer:

15,120

Step-by-step explanation:

The computation of the number of different routes are possible is shown below:

Since there are 5 locations for deliveries among the 9 locations that left

So, the number of routes that could be possible is

$= ^9P_5\\\\= \frac{9!}{9-5!} \\\\= \frac{9!}{4!} \\\\= \frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}\\\\= 15,120$

5 0

2 years ago