Given that ∠ABC ≅ ∠DBE, which statement must be true?

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

The volume of water in a tank is modeled by the function y = 1.15x3 – 0.1x2 + 2 where x is the number of minutes after the fauce

baherus [9]

Answer:

2

Step-by-step explanation:

in this context the y intercept represents the volume of water already in the tank

8 0

2 years ago

Read 2 more answers

Round 21,253 to the nearest hundred.

Mazyrski [523]

Answer:

Hey there! The answer to your question is 21,300

Step-by-step explanation:

If we look at the 253 it is most close to 300 not 200 so the answer will be 21,300

Hope that helps!

By: xBrainly

6 0

2 years ago

A biologist took a count of the number of fish in a particular lake and recounted the lake’s population of fish each of the next

oksano4ka [1.4K]

The <em>correct answer</em> is:

P(x) = 13x² – 10x + 350; 1,102 fish

Explanation:

I plotted this data in a graphing calculator. The screenshot is attached. You can see the curve of the parabola in the graph. I then ran the quadratic regression; again, the screenshot is attached. It lists the values of a, b and c for the equation y = ax²+bx+c.

8 0

2 years ago

Read 2 more answers

A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The

Oksanka [162]

Answer:

a) $[-0.134,0.034]$

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

$s_p^2 = \frac{ (n_1 -1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ ,

where $n_1 = 40, n_2 = 30, s_1 = 0.16, s_2 = 0.19$ .

The mean difference $\mu_1 - \mu_2 = 10.85 - 10.90 = -0.05$ .

The confidence interval is

$(\mu_1 - \mu_2) \pm t_{n_1+n_2-2,\alpha/2} \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} = (-0.05) \pm t_{68,0.025} \sqrt{\frac{0.03}{40} + \frac{0.03}{30}}$

$= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]$

b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.

c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.

6 0

2 years ago