The polygons below are similar. Find the value of y. (2 points)

A battery with 20% of its full capacity is connected to a charger. Every minute that passes, an additional 5% of its capacity is

Masja [62]

Answer:

Step-by-step explanation:

y = 5x + 20

Start at (0, 20).

Then plot a point at (1, 25).

The line should be going through points (2, 30), (3, 35), (4, 40), (5, 45), etc.

For every time the x number goes up, the y number goes up 5 times for the 5%.

3 0

1 year ago

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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

This graph models the exponential growth of the strain of bacteria. What does the ordered pair (15, 8) represent?

yan [13]

The answer is C.(There are 8 bacteria cells after 15 minutes.)

4 0

2 years ago

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An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The standard error of the difference (of the means) for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%).

If this interval contains the zero, we can say there is no significant difference.

Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples.

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.