Well this is very hard and I need help passing do anybody know the answer to this question

I'm am very confused with this question.

stich3 [128]

Answer:

103°

Step-by-step explanation:

This is a cyclic quadrilateral (all vertices on the same circle)

Opposite angles are supplementary:

∠A + ∠C = 180°
∠B + ∠D = 180°

Find x:

x + 5 + x - 7 = 180°
2x - 2 = 180°
2x = 182°
x = 91°

Find D:

∠D = x - 14°
∠D = 91° - 14° = 77°

Find B:

∠B = 180° - 77° = 103°

6 0

3 years ago

Consider independently rolling two fair dice, one red and the other green. Let A be the event that the red die shows 3 dots, B b

max2010maxim [7]

Answer:

A & B are independent events. A & C, B & C are not independent events.

Step-by-step explanation:

Independent Events are the events, whose occurrences are completely unrelated to each other.

Event A 'Red dice getting 3' & Event B 'Green die getting 4' ; are unrelated to each other. The numbers on red dice & on green dice are not dependent on each other.

Event C 'Red & Green dice numbers showing sum = 7' is dependent on the numbers coming on red & green dice , i.e events A & events B respectively. So, event C is dependent on both events A & B.

4 0

3 years ago

Fifty metal spheres, each with a radius of 2 cm, are melted. The melted solution is poured into a box. What is the volume of all

bazaltina [42]

Answer:

<h2>The first option is correct 1675.5 cm^3</h2>

Step-by-step explanation:

Step one:

we are given the radius as

r=2cm, and the number of spheres is 50 in numbers

let us find the volume of one sphere and multiply the area by 50 to get the volume of the 50 spheres

Step two:

$volume=4/3 \pi r^3$

$volume=4/3*3.142 *2^3\\\\volume=100.5/3\\\\volume=33.5$

The volume of one sphere is 33.5in^3

Hence the volume of the 50 spheres will be

=33.5*50

=1675.7cm^3 approx.

6 0

3 years ago

Suppose the time to complete a 200-meter backstroke swim for female competitive swimmers is normally distributed with a mean μ =

Natali [406]

Answer: The proportion that will qualify is 0.0314 or 3.14%.

First, we need to find the z-score of a time of 128 seconds. To do this, we find the difference of the mean and score and divide by the standard deviation.

(128 - 141) / 7 = -1.86

Now, use a standard normal distribution table to determine the percent below a z-score of -1.86. That value is 0.0314 or 3.14%.

6 0

3 years ago

A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F

Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let X = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

$P(X)=p=\frac{10}{18}=0.556$

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of X is given by:

$P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...$

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

$P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531$

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

$=\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959$

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747$

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0

3 years ago