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3 years ago

17. You have $75 in your bank account. Each week you plan to deposit $8 from your allowance and $10 from your paycheck

Mathematics

1 answer:

Vanyuwa [196]3 years ago

6 0

Answer:

60 weeks

Step-by-step explanation:

195 = 75 + (10 + 8)w

195 - 75 = (10 + 8)w

120 = (10 + 8)w

120 = (2)w

120 ÷ 2 = w

60 = w

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6. (4.11.2) A 500-page book contains 250 sheets of paper. The thickness of the paper used to manufacture the book has mean 0.08

Maksim231197 [3]

Answer:

10.38% probability that a randomly chosen book is more than 20.2 mm thick

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

For sums of n elements, the mean is $n\mu$ and the standard deviation is $s = \sqrt{n}\sigma$

In this problem, we have that:

$\mu = 250*0.08 = 20, \sigma = \sqrt{250}*0.01 = 0.1581$

What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)

This is 1 subtracted by the pvalue of Z when X = 20.2. So

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

By the Central Limit Theorem

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

$Z = \frac{20.2 - 20}{0.1581}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962

1 - 0.8962 = 0.1038

10.38% probability that a randomly chosen book is more than 20.2 mm thick

3 0

2 years ago

HELP FAST PLEASE!!

stira [4]

I think it is B. and C.
secx = 3 and tanx = 3
Hope it helps

3 0

3 years ago

Read 2 more answers

5.2.14. For the negative binomial pdf p (k; p, r) = k+r−1 (1 − p)kpr, find the maximum likelihood k estimator for p if r is know

Volgvan

Answer:

$\hat p = \frac{r}{\bar x +r}$

Step-by-step explanation:

A negative binomial random variable "is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution, this distribution is known as the Pascal distribution".

And the probability mass function is given by:

$P(X=x) = (x+r-1 C k)p^r (1-p)^{x}$

Where r represent the number successes after the k failures and p is the probability of a success on any given trial.

Solution to the problem

For this case the likehoof function is given by:

$L(\theta , x_i) = \prod_{i=1}^n f(\theta ,x_i)$

If we replace the mass function we got:

$L(p, x_i) = \prod_{i=1}^n (x_i +r-1 C k) p^r (1-p)^{x_i}$

When we take the derivate of the likehood function we got:

$l(p,x_i) = \sum_{i=1}^n [log (x_i +r-1 C k) + r log(p) + x_i log(1-p)]$

And in order to estimate the likehood estimator for p we need to take the derivate from the last expression and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\frac{x_i}{1-p}$

And we can separete the sum and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}$

Now we need to find the critical point setting equal to zero this derivate and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}=0$

$\sum_{i=1}^n \frac{r}{p} =\sum_{i=1}^n \frac{x_i}{1-p}$

For the left and right part of the expression we just have this using the properties for a sum and taking in count that p is a fixed value:

$\frac{nr}{p}= \frac{\sum_{i=1}^n x_i}{1-p}$

Now we need to solve the value of $\hat p$ from the last equation like this:

$nr(1-p) = p \sum_{i=1}^n x_i$

$nr -nrp =p \sum_{i=1}^n x_i$

$p \sum_{i=1}^n x_i +nrp = nr$

$p[\sum_{i=1}^n x_i +nr]= nr$

And if we solve for $\hat p$ we got:

$\hat p = \frac{nr}{\sum_{i=1}^n x_i +nr}$

And if we divide numerator and denominator by n we got:

$\hat p = \frac{r}{\bar x +r}$

Since $\bar x = \frac{\sum_{i=1}^n x_i}{n}$

4 0

3 years ago

Find the total service area of this cone. leave your answer in terms of pi. r=10in h=24in

daser333 [38]

Answer:

36pie inch²

Step-by-step explanation:

Answer attached

Hope it helps :)

5 0

2 years ago

Which linear function represents the line given by the point-slope equation y + 7 = -2/3 (x + 6)?

PolarNik [594]

Got this from someone else but it answers your question

O f(x) = -x-1
Step-by-step explanation:
We want to isolate the 'y' variable, so we will subtract 7 from both sides.
y = x + 6 - 7
y = x -1

also can u please give me brainliest

6 0

2 years ago