Please help with this question!

If 120% is £48, what is 100%

NeX [460]

120% = 120/100 = 1.2

48/1.2 = 40

Answer: 40.

4 0

3 years ago

Write two expressions that are equivalent to the expression below.<br> (9-0) + (9-c) + (9-c) + (9-C)

Nikolay [14]

Step-by-step explanation:

$4(9 - c) \\ 36 - 4c \\ an \: example \: of \: two \: equivalent \\ expression \: includes \: \\ the \: general \: experssion \: \\ and \: then \: the \: one \: that \: \\ the \: one \: that \: is \: factured$

4 0

3 years ago

A water sprinkler sends water out in a circular pattern. How many feet away from the sprinkler can it spread water if the area f

myrzilka [38]

Answer:

The sprinkler can spread water 18 feet away.

Step-by-step explanation:

We are given the following in the question:

Area formed by watering pattern = 1,017.36 square feet

We have to find the how far the sprinkler spread the water.

The sprinkler covers a circular area. We need to find the radius of this circular area to find the how far the sprinkler spread the water.

Area of circle =

$A = \pi r^2$

where r is the radius of the circle.

Putting values, we get,

$1017.36 = 3.14 r^2\\\\r^2 = \dfrac{1017.36}{3.14}\\\\r^2 = 324\\r = \sqrt{324} = 18\text{ feet}$

Thus, the sprinkler can spread water 18 feet away.

6 0

3 years ago

Read 2 more answers

A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly

Keith_Richards [23]

Answer:

0.7208 = 72.08% probability that this whole shipment will be accepted.

Step-by-step explanation:

For each tablet, there are only two possible outcomes. Either it meets the required specifications, or it does not. The probability of a tablet meeting the required specifications is independent of any other tablet, which means that the binomial probability distribution is used to solve this question.

Binomial probability 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.

4% rate of defects

This means that $p = 0.04$

26 tablets

This means that $n = 26$

What is the probability that this whole shipment will be accepted?

Probability that at most one tablet does not meet the specifications, which is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Thus

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

$P(X = 0) = C_{26,0}.(0.04)^{0}.(0.96)^{26} = 0.3460$

$P(X = 1) = C_{26,1}.(0.04)^{1}.(0.96)^{25} = 0.3748$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3460 + 0.3748 = 0.7208$

0.7208 = 72.08% probability that this whole shipment will be accepted.

3 0

3 years ago

A given population proportion is .25. What is the probability of getting each of the following sample proportions

anyanavicka [17]

This question is incomplete, the complete question is;

A given population proportion is .25. What is the probability of getting each of the following sample proportions

a) n = 110 and = p^ ≤ 0.21, prob = ?

b) n = 33 and p^ > 0.24, prob = ?

Round all z values to 2 decimal places. Round all intermediate calculation and answers to 4 decimal places.)

Answer:

a) the probability of getting the sample proportion is 0.1660

b) the probability of getting the sample proportion is 0.5517

Step-by-step explanation:

Given the data in the questions

a)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 110

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 110 ) √( 0.1875 / 110 ) = 0.0413

Now, P( p^ ≤ 0.21 )

= P[ (( p^ - μ ) /S.D) < (( 0.21 - μ ) / S.D)

= P[ Z < ( 0.21 - 0.25 ) / 0.0413)

= P[ Z < -0.04 / 0.0413]

= P[ Z < -0.97 ]

from z-score table

P( X ≤ 0.21 ) = 0.1660

Therefore, the probability of getting the sample proportion is 0.1660

b)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 33

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 33 ) = √( 0.1875 / 33 ) = 0.0754

Now, P( p^ > 0.24 )

= P[ (( p^ - μ ) /S.D) > (( 0.24 - μ ) / S.D)

= P[ Z > ( 0.24 - 0.25 ) / 0.0754 )

= P[ Z > -0.01 / 0.0754 ]

= P[ Z > -0.13 ]

= 1 - P[ Z < -0.13 ]

from z-score table

{P[ Z < -0.13 ] = 0.4483}

1 - 0.4483

P( p^ > 0.24 ) = 0.5517

Therefore, the probability of getting the sample proportion is 0.5517

6 0

3 years ago