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3 years ago

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Mathematics

2 answers:

Zepler [3.9K]3 years ago

6 0

Answer:

In step 2, she substituted the x values for y and the y values for x.

Step-by-step explanation:

just did it

Pavlova-9 [17]3 years ago

5 0

Answer:

step two is wrong

Step-by-step explanation:

You might be interested in

If he is correct, what is the probability that the mean of a sample of 68 computers would differ from the population mean by les

elena-14-01-66 [18.8K]

Complete Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months with a standard deviation of 10 months if he is correct. what is the probability that the mean of a sample of 68 computers would differ from the population mean by less than 2.08 months? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad

Answer:

$P(-1.72$

Step-by-step explanation:

From the question we are told that:

Population mean \mu=91

Sample Mean \=x =2.08

Standard Deviation \sigma=10

Sample size n=68

Generally the Probability that The sample mean would differ from the population mean

P(|\=x-\mu|<2.08)

From Table

$P(|\=x-\mu|$

T Test

$Z=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }$

$Z=\frac{2.08}{\frac{10}{\sqrt{68} } }$

$Z=1.72$

$P(|\=x-\mu|$

$P(-1.72$

Therefore From Table

$P(-1.72$

5 0

3 years ago

2. The store bought a pair of shoes for $60, and sold

seropon [69]

Answer:

33.3%

Step-by-step explanation:

Given data

Cost of shoes= $60

Selling price = $100

% markup = (selling-cost/cost)*100

% markup = (100-75/75)*100

% markup = (25/75)*100

% markup = (0.33333333333)*100

% markup =33.3%

Hence the markup is 33.3%

3 0

3 years ago

Find the balance in the account. $2,500 principal earning 4%, compounded quarterly, after 4 years

Helen [10]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2500\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &4 \end{cases} \\\\\\ A=2500\left(1+\frac{0.04}{4}\right)^{4\cdot 4}\implies A=2500(1.01)^{16}\implies A\approx 2931.4466123$

3 0

4 years ago

Can some help me with this question

Luden [163]

To find the equation of this line in slope-intercept form (y = mx + b, where m is its slope and b is its y-intercept), we naturally need the slope and the y-intercept. We can see that the line intersects the y-axis at the point (0, 4) so our y-intercept is 4, and the line rises 4 along the y-axis for every 2 it runs along the x-axis, so its slope is 4/2 = 2. With this in mind, we can write the line's equation as

y = 2x + 4

7 0

3 years ago

You are running a race. The probability that you win is 3/5. What is the probability that you lose at most 2 out of your 6 races

Veronika [31]

Answer:

The probability that you lose at most 2 out of your 6 races is 0.54432.

Step-by-step explanation:

We are given that you are running a race. The probability that you win is 3/5.

There are total of 6 races.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 6 races

r = number of success = at most 2 lost

p = probability of success which in our question is probability that

you lose a race = 1 - (3/5) = 2/5 or 0.4

Let X = Number of races lost

So, X ~ Binom(n = 6, p = 0.40)

Now, the probability that you lose at most 2 out of your 6 races is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{6}{0} \times 0.40^{0} \times (1-0.40)^{6-0} + \binom{6}{1} \times 0.40^{1} \times (1-0.40)^{6-1} + \binom{6}{2} \times 0.40^{2} \times (1-0.40)^{6-2}$

= $1 \times1 \times 0.60^{6} + 6 \times 0.40^{1} \times 0.60^{5} +15\times 0.40^{2} \times 0.60^{4}$

= 0.54432

7 0

3 years ago