Is the following number rational or irrational?

A sail is in the form of a right triangle that is two times as high as it is wide. The sail is made from 4 square meters of mate

Ivahew [28]

<h3>Answer:</h3>

4 m

<h3>Explanation:</h3>

If we let h represent the height of the sail, then its width is h/2. The area of the sail is given by ...

... A = (1/2)wh

Filling in the given information, we have

... 4 m² = (1/2)(h/2)h

... 16 m² = h² . . . . . . multiply by 4

... 4 m = h . . . . . . . . take the square root

4 0

3 years ago

According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what i

Ierofanga [76]

Answer:

a) $P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

b) $P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

c) For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

And replacing we got:

$P(X\geq 8)=0.0000779$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n=10 ,p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Let X the random variable "number of women that have never been married" , on this case we now that the distribution of the random variable is:

$X \sim Binom(n=10, p=0.2)$

Part a

We want to find this probability:

$P(X=2)$

And using the probability mass function we got:

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

Part b

For this case we want this probability:

$P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

We can find the individual probabilities and we got:

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

Part c

For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

3 0

3 years ago

If the factors of quadratic function g are (x − 7) and (x + 3), what are the zeros of function g?

MArishka [77]

Answer: B. -3 and 7

Step-by-step explanation:

To find the zeros, just move the constants to the other side,

x - 7 ⇒ x = 7

x + 3 ⇒ x = -3

If the number is moved to the other side its sign becomes opposite

Hope it helps!

8 0

2 years ago

The medication tablets contain 3.5 grains each. The physician's orders state that the patient is to receive 14 grains of medicat

Lubov Fominskaja [6]

14 / 3.5 = 4.....4 tablets should be administered

7 0

3 years ago

I have 47 nickels and dimes worth $4.20. How many of each type of coin do I have?

soldi70 [24.7K]

Answer:

37 dimes and 10 nickels

Step-by-step explanation:

let d = # dimes

let n = # nickels

we can set up a system of equations:

n + d = 47

.05n + .10d = 4.2

if we solve the first equation for 'n' we get:

n = 47-d

now we can substitute this in for 'n' in the second equation:

.05(47-d) + .10d = 4.2

2.35 - .05d + .10d = 4.2

2.35 + .05d = 4.2

subtract 2.35 from each side to get:

.05d = 1.85

d = 1.85÷.05

d = 37

if d+n = 47 and d=37 then n = 10

Check:

.05(10) + .1(37) should equal 4.2

.50 + 3.7 = 4.2 [It Checks Out]

6 0

2 years ago