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2 years ago

15

The data below represents the scores in a golf tournament. If the mean is 70 with a standard deviation of 4.9, circle all values

with a z-score greater than 0.5
{63, 64, 65, 67, 70, 71, 73, 73, 76, 78}

1 answer:

ololo11 [35]2 years ago

5 0

Answer:

70 is the anwer I got I am so sorry if that was wrong but it should be right

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Kira received a $1600 bonus. She decided to invest it in a 3-year certificate of deposit (CD) with an annual interest rate of 1.

erma4kov [3.2K]

Answer:

1. $1666.17

2. $666.17

Step-by-step explanation:

The bonus of $1600 that Kira gets is invested for 3 years at an annual interest of 1.36% compounded annually.

Therefore, after 3 years Kira will have in her account

$1600(1+\frac{1.36}{100} )^{3} =1600(1.0136)^{3}= 1666.17$ dollars (Answer)

Therefore, the amount of interest earned by Kira's investment after 3 years will be

$( 1666.17 - 1600 ) = $666.17 ( Answer )

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3 years ago

Write in index form using a unit fraction : 6√19

Fynjy0 [20]

$6\sqrt{19} =\sqrt{36}\sqrt{19}$

$=\sqrt{684}$

$=684^{1/2}$

Hence, the index form of $6\sqrt{19 } =684^{1/2}$ .

6 0

3 years ago

I'll give 30 points and brainlist

Kamila [148]

Answer:

100 degrees

Step-by-step explanation:

Arc QM = Arc QN = 130°

Arc MN = 360 - Arc QM - Arc QN

= 360 - 2(130)

= 360 - 260

= 100°

7 0

4 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

Which of the following transformations fit the definition of isometry?

Gennadij [26K]

Answer: Rotations, reflections, translations (A, C, and E)

Imagine you had a camera aimed at a triangular figure on a piece of paper. If you rotate the camera, then the image of the triangle appears to rotate. In reality it's the other way around. What this means is that the triangle is not changing at all. It keeps the same size, shape, area, perimeter, etc. This applies to when the camera pans left or right, ie shifts from side to side. The triangle will translate but again the triangle isn't changing at all. It's merely an illusion. Reflections are the same way. Imagine having a piece of glass or a mirror that reflects the image which is an identical copy; although everything is flipped.

Dilations are not isometries because the image is a different size then the pre-image. The same shape is maintained though. Note: the scale factor must be some number other than 1.

another note: "isometry" breaks down into "iso+metry" with "iso" meaning "same" or "equal", and "metry" meaning "measure". So if you had 2 identical yard sticks, then they are isometrical or equal in length.

4 0

3 years ago