Write 1.33 as a percentage

STatiana [176]

For the first one, you're answer would be AC. since C is also a right angle, and angle A is hypnotenuse to C

3 0

3 years ago

In a sample of n = 6 scores, 5 of the scores are each above the mean by one point. Where is the 6th score located relative to th

den301095 [7]

The mean of a dataset is the sum of all data elements divided by the count of the elements.

The location of the 6th score relative to the mean is 5 points below the mean

Let:

$\bar x \to$ Mean

$a \to$ 5 scores

$b \to$ 6th scores

Given that:

$n = 6$

The 5 scores that are 1 above the mean implies that:

$a = \bar x + 1$

The mean of a dataset is calculated using:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x =\frac{5a + b}{6}$

$\bar x =\frac{5(\bar x + 1) + b}{6}$

Open brackets

$\bar x =\frac{5\bar x + 5 + b}{6}$

Multiply both sides by 6

$6\bar x =5\bar x + 5 + b$

Make b the subject

$b = 6\bar x -5\bar x - 5$

$b = \bar x - 5$

This means that the 6th score is 5 points below the mean

Read more about mean at:

brainly.com/question/17060266

3 0

2 years ago

An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. (a) Use a bin

Advocard [28]

Answer:

a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

$\mu = np$

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that $p = \frac{1}{2500} = 0.0004$

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is $P(X = 4)$ when $n = 7000$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558$

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

$\mu = np = 7000*0.0004 = 2.8$ . So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557$

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

6 0

3 years ago

A rectangle has a width of 8 centimeters and an area of 160 square centimeters a similar rectangle has an area of 250 square cen

Bezzdna [24]

I'm a bit confused and there is not a lot of information given to this.

From what I know, the dimensions of the larger rectangle can be found from length x width = 250 square centimetres.

Any multiplication problem that equals 250 could work out for length x width.

8 0

3 years ago

HELPP ASAP I WILL GIVE 14 POINTS

jek_recluse [69]

Given:

A box-and-whisker plot of data set.

To find:

The percentage of the data values that are greater than 80.

Solution:

From the given box-and-whisker plot, it is clear that:

Minimum value = 8

First quartile = 60

Median = 68

Third quartile = 80

Maximum value = 92

We know that the 25% the data value are greater than or equal to third quartile because the third quartile divides the data in 75% to 25% and 80 is the third quartile.

Therefore, about 25% of the data values that are greater than 80.

5 0

2 years ago