Which function has an inverse that ia also a function

You are making a picture frame of uniform width for a picture that is to be displayed at the local museum. The picture is 3.25 f

SCORPION-xisa [38]

$x-\ width\ preferred\\\\
Area=width*length\\\\
length=3,25feet\\
Area=15feet^2\\\\
15=3,25*x\ \ \ \ | divide\ by\ 3,25\\\\x=4,61feet\\\\It\ should\ be\ 4,61\ feet \ long.
$

8 0

3 years ago

A market research company conducted a survey to find the level of affluence in a city. They defined the category "affluence" for

malfutka [58]

Answer:

A 95% confidence interval for this population proportion is [0.081, 0.159].

Step-by-step explanation:

We are given that a market research company conducted a survey to find the level of affluence in a city.

Out of 267 persons who replied to their survey, 32 are considered affluent.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who are considered affluent = $\frac{32}{267}$ = 0.12

n = sample of persons = 267

p = population proportion

Here for constructing a 95% confidence interval we have used One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ , $0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ ]

= [0.081, 0.159]

Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].

4 0

3 years ago

your friend drinks 2/3 of a bottle of water you drink 5/7 of a bottle of water find the difference of amounts of water left in e

horsena [70]

Well you would convert the fractions to fractions with the same denominator which would make it 14/21 for the first friend and 15/21 for the second friend. But the question is asking how much is left so you have to subtract what is left in each bottle. The first friend has 7/21 left and the second friend has 6/21 left. So, the answer is 1/21.

7 0

3 years ago

Guys help my little sister with these questions:)))

sveta [45]

Answer:

Card A: the eiffel tower is 20,313 taller than the pyramid of khafria

Card B: Josh site had 615 more than diego

Card C: 15,405 more to reach there goal

Step-by-step explanation:

Tell your sister i said your welcome

6 0

3 years ago

Read 2 more answers

QUESTION THREE (30 MARKS) 3.1 The mass of a standard loaf of white bread is, by law meant to be 700g with a population standard

kiruha [24]

Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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Mean of 700g means that $\mu = 700$
Standard deviation of 21g means that $\sigma = 21$
Sample of 64, thus $n = 64$
For the sampling distribution of the sample mean, the standard deviation is of $s = \frac{21}{\sqrt{64}} = \frac{21}{8} = 2.625$