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Bas_tet [7]
3 years ago
15

445 randomly selected light bulbs were tested in a laboratory, 316 lasted more than 500 30) hours. find a point estimate of the

proportion of all light bulbs that last more than 500 hours.
Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0
Total number of bulbs tested, N = 445
Number of bulbs that lasted more than 500 hrs, x = 316

Point estimate of bulbs that lasted more than 500 hrs = x/N = 316/445 ≈ 0.71 = 0.71*100 = 71%
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Determine which of the following sets of ordered pairs represents a function
Illusion [34]

Answer:

B.

Step-by-step explanation:

Every answer besides B. has an input that repeats. A function contains inputs that have exactly one output. This means x values cannot have more than one y value.

8 0
3 years ago
The radius of a circular merry-go-round is 3 meters. which measurement is closest to the area of the merry-go-round in square me
djyliett [7]

Answer:

area of the merry go round =πr²

=22/7×3²

=28.28

answer is option D

4 0
2 years ago
Angle of a triangle measures 115° the other two angles are in the ratio of 4:9 what are the measures of those two angles
nasty-shy [4]

Answer:

20, 45

Step-by-step explanation:

A triangle has 180 degrees interior angle measure. One angle measure 115 so that means the other two angles  must add up to 65.

The remaning angles form a ratio of 4:9. This means we must split 65 into a ratio of. 4:9

A ratio partition parts of something. We can add the ratio intergers to find it full length.

4+9=13.

Divide 65/13=5.

Multiply 5x4 and 5x9 serpately.

We get 20 and 45

6 0
2 years ago
Two problems here I need solved! I need every step, so please have that with your answers!!
Free_Kalibri [48]
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
is not in the domain of the given equation.

It is an extraneous solution.

\therefore \: x = - 1
is the only solution.

QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
7 0
3 years ago
When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F
aleksandr82 [10.1K]

Let x be the unknown number. So, three times that number means 3x, and the square of the number is x^2

We have to sum 528 and three times the number, so we have 528+3x

Then, we have to subtract this number from x^2, so we have

x^2-(3x+528)

The result is 120, so the equation is

x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0

This is a quadratic equation, i.e. an equation like ax^2+bx+c=0. These equation can be solved - assuming they have a solution - with the following formula

x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

If you plug the values from your equation, you have

x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}

So, the two solutions would be

x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27

x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24

But we know that x is positive, so we only accept the solution x = 27

6 0
3 years ago
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