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1 year ago

15

The average age three people running for election is 42. A fourth person joins the race and the average drops to 40. What is the

fourth person's age?

1 answer:

bogdanovich [222]1 year ago

5 0

Let $a_1,\ldots,a_4$ denote the ages of the 4 candidates. Then

$\dfrac{a_1+a_2+a_3+a_4}4=40$

$a_1+a_2+a_3+a_4=160$

$\dfrac{a_1+a_2+a_3}3+\dfrac{a_4}3=\dfrac{160}3$

The average age of the first 3 candidates is 42, so

$\dfrac{a_4}3=\dfrac{160}3-42$

$\implies\boxed{a_4=160-3\cdot42=34}$

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If 3.0 grams of strontium-90 in a rock sample remained in 1989, approximately how many grams of strontium-90 were present in the

kirill [66]

Answer:

<em>11.29 grams </em><em>of strontium-90 were present in the original rock sample in 1933.</em>

Step-by-step explanation:

Strontium has a half-life of 28.8 years. Therefore,

1989 - 1933 = 56 years.

56 / 28.8 = 1.94 half-lives

Thus, the quantity of the radioisotope remaining will double for each half-life elapsed. moving backwards in time.

Therefore, moving backwards in time by 1.94 half-lives, the quantity remaining will double by 1.94 times.

Thus, the amount remaining in 1933 is

3.0 × (1.94)² = <em>11.29 grams</em>

5 0

2 years ago

Select all of the answers below that are equal to B = {John, Paul, George, Ringo, Pete, Stuart}

UNO [17]

Answer:

The correct option is 5) {Paul, Ringo, Pete, John, George, Stuart}.

Step-by-step explanation:

Consider the provided sets:

B = {John, Paul, George, Ringo, Pete, Stuart}

Two sets are equal if all the elements of Sets are same.

Set B has the elements: John, Paul, George, Ringo, Pete and Stuart

Now consider the provided options of sets.

From the provided options of set only option 4 has all the elements of set B but the order is different.

Thus, the correct option is 5) {Paul, Ringo, Pete, John, George, Stuart}.

8 0

1 year ago

Triangle G F E is cut by line segment H J. Line segment H J goes from side G E to F E. The length of G F is 4 x minus 4, the len

alexira [117]

Answer:

HJ = 8 JE = 4

Step-by-step explanation:

it is given that H is the midpoint of GE and J is the midpoint of FE. According to the midpoint theorem the line segment connecting the midpoint of two sides is parallel to the three side and its length is half of the third side. since JH is connecting the midpoints.

HJ= 1/2 (GF)

x + 3 = 1/2 (4x - 4)

x + 3 = 2x - 2

x = 5

^ Thus meaning the value of x is 5.

Now you just fill into your equations:

HJ = x + 3 = (5) + 3 = 8

JE = x - 1 = (5) - 1 = 4

Therefore, HJ = 8; JE = 4.

5 0

1 year ago

Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).

Elan Coil [88]

Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

$a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$

Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

Then, triple integral is:

$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

Calculating:

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

$\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx$

$\frac{1}{45} \int\limits^5_0 {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4} } \, dx$

$\frac{1}{45} [30375-60750+118462.5-39150]$

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$ = 1087.5

<u>The volume of the tetrahedon is 1087.5 cubic units.</u>

3 0

1 year ago

Jordan used the distributive property to write an expression that is equivalent to 6c – 48. 6c - 48 is equivalent to 6(c - 48) I

Katyanochek1 [597]

It’s incorrect.

The correct answer is 6(c - 8)

8 0

2 years ago

Read 2 more answers