Write 11,760,825 in word form and expanded form

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0

2 years ago

Triangle L N M is shown. Angle L N M is a right angle. An altitude is drawn from point N to point O on side L M, forming a right

zaharov [31]

Answer:

Therefore the value of k = 6.

Step-by-step explanation:

Given:

LN = m

NM = l

OM = k

NO = 4

LO = 8

LM = 8 + k and

Δ LNM ,Δ LON and Δ MON are right Triangle.

To Find :

Om = k = ?

Solution:

In Right angle Triangle By Pythagoras Theorem we have,

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

So, In Right angle Triangle Δ LON we have,

LN² = ON² + OL²

m² = 4² + 8²

m² = 80 ............( 1 )

Now in Right angle Triangle Δ MON we have,

MN² = ON² + MO²

l² = 4² + k² ....................( 2 )

Now In Right angle Triangle Δ LNM we have,

LM² = LN² + MN²

(8 + k)² = m² + l² .................( 3 )

Substituting equation 1 and equation 2 in equation 3

(8+k)² = 80 + 4² + k²

Applying (A+B)² = A² +2AB + B² we get

$64 + 16k+k^{2} = 80+ 16 +k^{2} \\\\16k=96\\\\\therefore k=\frac{96}{16} \\\\\therefore k=6$

Therefore the value of k = 6.

4 0

1 year ago

Read 2 more answers

Complete the steps for solving 7 = –2x2 + 10x. Factor out of the variable terms. inside the parentheses and on the left side of

Mamont248 [21]

we have

$7=-2x^{2} +10x$

Factor the leading coefficient

$7=-2(x^{2} -5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$7-12.50=-2(x^{2} -5x+2.5^{2})$

$-5.50=-2(x^{2} -5x+2.5^{2})$

Divide both sides by $-2$

$2.75=(x^{2} -5x+2.5^{2})$

Rewrite as perfect squares

$2.75=(x-2.5)^{2}$

Taking the square roots of both sides (square root property of equality)

$x-2.5=(+/-)\sqrt{2.75}$

Remember that

$\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}$

$x-2.5=(+/-)\frac{\sqrt{11}}{2}$

$x=2.5(+/-)\frac{\sqrt{11}}{2}$

$x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}$

$x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}$

the answer is

The solutions are

$x=\frac{5+\sqrt{11}}{2}$

$x=\frac{5-\sqrt{11}}{2}$

5 0

1 year ago

Read 2 more answers

The number of pupils in a school increases at the rate of of 5% per annum. When the school opened there were 200 pupils on the r

Rashid [163]

Answer:

I have no idea what the answer is

8 0

2 years ago

The mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation i

olga_2 [115]

Answer:

The z-score (value of z) for an income of $1,100 is 1.

Step-by-step explanation:

We are given that the mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation is $100.

Let X = group of weekly incomes of a large group of executives

So, X ~ N( $\mu=1,000 ,\sigma^{2} = 100^{2}$ )

The z-score probability distribution for a normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean income = $1,000

$\sigma$ = standard deviation = $100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, we are given an income of $1,100 for which we have to find the z-score (value of z);

So, z-score is given by = $\frac{X-\mu}{\sigma}$ = $\frac{1,100-1,000}{100}$ = 1

Hence, the z-score (value of z) for an income of $1,100 is 1.

4 0

1 year ago