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Ira Lisetskai [31]
3 years ago
11

3x + 8y = 12 2x + 2y = 3 Part A: Create an equivalent system of equations by replacing one equation with the sum of that equatio

n and a multiple of the other. Show the steps to do this. (6 points) Part B: Show that the equivalent system has the same solution as the original system of equations. (4 points)
Mathematics
1 answer:
love history [14]3 years ago
7 0
Ok here is what I think.
Let us first number these statements, as #1, and #2.
First statement: 3x + 8y = 12 (1)
Second Statement: 2x + 2y = 3 (2)
Now, we can work from this.
We want to make one of the equations be equal to 0 so that at the end when we check they can be equal to each other.
Let us use 4.
3x+8y=12      1-8x-8y=-12    2
This gives us:-5x = 0          
Now we should try and isolate x so we can substitute it into one of the equations.
We have -5x=0
and x=0
3(0)+8y=12
8y=12
y=12/8
y=3/2
Plug in these new equations
y=3/2 and y=0 into any of the first equations
 3x+8y=12                3(0)+8(3/2)= 12       8(3/2)=12                 4(3)=12                    12=12                           
Now we know it works, thats our check^^
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Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a
Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
3 years ago
HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!
mrs_skeptik [129]

Answer:

See below.

Step-by-step explanation:

Just make a line that doesn't have a pattern like 2, 4, 6, 8 or like 6, 12, 18, 24.

99.9% sure just spamming the keyboard with random numbers will work.

5 0
3 years ago
Tank A has a capacity of 9.5 gallons. 6 1\3 gallons of the tank"s water are poured out. How many gallons of water are left in th
dezoksy [38]
Since 1/3 can't be turned into a decimal, change .5 into 1/2. In order to subtract, the fractions need a common denominator. The least common denominator is 6. So 1/3 turns into 2/6, and 1/2 turns into 3/6. Now you need to take 9 3/6 minus 6 2/6. The answer is 3 1/6 gallons of water.

3 0
3 years ago
1. The Senior Class of Great Ridge High School and Ann Arbor High School are planning a trip to Six Flags Theme Park. Great Ridg
Zanzabum

Answer:

A)14v + 16b = 1152

10v + 13b = 925

B)Number of students in a van = 8

Number of students in each bus = 65

C) Elimination method was easier.

Step-by-step explanation:

Let v be the number of students in a van

Let b be number of students in a bus

A) We are told the first school can fill 14 vans and 16 buses with 1152 students

Thus;

14v + 16b = 1152 - - - - - (eq 1)

Also,we are told that the second school can fill 10 vans and 13 buses with 925 students.

Thus;

10v + 13b = 925 - - - - (eq 2)

B) To solve this we will use elimination method.

Multiply eq (1) by 5 and multiply eq (2) by 7 to give;

70v + 80b = 5760 - - - - eq(3)

70v + 91b = 6475 - - - - - eq(4)

Subtract eq(3) from eq(4) to give;

91b - 80b = 6475 - 5760

11b = 715

b = 715/11

b = 65

Put 65 for b in eq(3) to give;

70v + 80(65) = 5760

70v = 5760 - 80(65)

70v = 560

v = 560/70

v = 8

Number of students in a van = 8

Number of students in each bus = 65

C) Elimination method was used because in both equations, we had different digits attached and so it would have been more tedious to use substitution method or graphical method. Thus, elimination method was more easier.

7 0
3 years ago
The volume of a sphere is 113.04 cubic feet. Find the radius. Use 3.14 for pi.
avanturin [10]
Volume is 4/3(3.14)(r)^2
the radius is 5.2 ft
4 0
3 years ago
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