3x + 15 2. Find the value of X

The average amount of money spent for lunch per person in the college cafeteria is $6.35 and the standard deviation is $2.31. Su

alexdok [17]

Answer:

(6.35, 5.34)

(6.35, 0.99)

0.10253

0.3984

Step-by-step explanation:

Given that :

μ = 6.35

σ = 2.31

n = 41

What is the distribution of X?X ~ N(,)

X :

Mean of distribution = μ= 6.35

The variance = σ^2 = 2.31^2 = 5.3361 = 5.34

X ~ N = (6.35, 5.34)

What is the distribution of x¯? x¯ ~ N(,)

The mean = μ = 6.35

The standard deviation of the mean = σ/sqrt(n) = 6.35/sqrt(41) = 0.9917 = 0.99

X ~N = (6.35, 0.99)

find the probability that this patron's lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757)

Obtain the standardized scores:

Z = (x - μ) / σ ; (6.1605 - 6.35) / 2.31 = - 0.082

P(Z < - 0.082) = 0.46732 (Z probability calculator)

(6.757 - 6.35) / 2.31 = 0.176

P(Z < 0.176) = 0.56985 (Z probability calculator)

P(Z < 0.176) -P(Z < - 0.082)

0.56985 - 0.46732 = 0.10253

For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757) ; n = 17

Obtain the standardized scores:

Z = (x - μ) / σ/sqrt(n) ; (6.1605 - 6.35) / 2.31/sqrt(17) = - 0.338

P(Z < - 0.338) = 0.36768 (Z probability calculator)

(6.757 - 6.35) / 2.31/sqrt(17) = 0.726

P(Z < 0.726) = 0.76608 (Z probability calculator)

P(Z < 0.726) -P(Z < - 0.338)

0.76608 - 0.36768 = 0.3984

8 0

3 years ago

In the diagram abc=adb=90, ad=p and dc=q. Use similar triangles to show that x2=pz plzz anyoneee

kramer

Answer:

By comparing the ratios of sides in similar triangles ΔABC and ΔADB,we can say that $x^{2} =pz$

Step-by-step explanation:

Given that ∠ABC=∠ADC, AD=p and DC=q.

Let us take compare Δ ABC and Δ ADB in the attached file , ∠A is common in both triangles

and given ∠ABC=∠ADB=90°

Hence using AA postulate, ΔABC ≈ ΔADB.

Now we will equate respective side ratios in both triangles.

$\frac{AB}{AC}= \frac{AD}{AB}=\frac{BD}{BC}$

Since we don't know BD , BC let us take first equality and plugin the variables given in respective sides.

$\frac{x}{z}= \frac{p}{x}$

Cross multiply

$x^{2}=pz$

Hence proved.

7 0

3 years ago

A rancher wants to cut a metal cylinder in half to use as a watering trough as shown below.

velikii [3]

Answer:

88.3 ft3 is the answer hope this helps!!

Step-by-step explanation:

3 0

3 years ago

Your company has offices in 2 cities. The Denver office has 137 employees with an average salary of $67,013. The Anchorage offic

Charra [1.4K]

Alright, lets get started.

The Denver office has 137 employees with an average salary of $67,013.

So the total salay of all employees at Denver office will be = $137*67013$

So the total salay of all employees at Denver office will be = $ 9180781

The Anchorage office has 25 employees with an average salary of $71,332.

So the total salary of employees at Anchorage office will be = $25 * 71332$

So the total salary of employees at Anchorage office will be = $ 1783300

So, complete total salary of employees at both offices will be = $9180781+1783300$

So, complete total salary of employees at both offices will be = $ 10964081

Total number of employees at both offices are = $137 + 25 = 162$

So, the average salary of employee across the company = $\frac{10964081}{162}$

So, the average salary of employee across the company = 67679.5

Rounding off,

So, the average salary of employee across the company = $ 67680 : Answer

Hope it will help :)

7 0

3 years ago

Read 2 more answers

Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

18 values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above 18 values for n are possible.

3 0

3 years ago