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2 years ago

P(A)=7/10 p(AorB)=173/200 what is p(B)?

1 answer:

4 0

If A and B are mutually exclusive
P(AorB) = P(A)+P(B)
173/200 = 7/10+ P(B)
173/200 - 7/10 = P(B)
173/200 - 140/200 = P(B)
33/200 = P(B)

Hope it helps.

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If mJI = (3x+2)°, mHLK = (15x-36)°, and m∠HML = (8x-1)°, find mHLK

Gala2k [10]

Answer:

The value of mHLK will be "(204)°".

Step-by-step explanation:

The given values are:

mJI = (3x+2)°

mHLK = (15x-36)°

and,

m∠HML = (8x-1)°

then,

mHLK = ?

Now,

By using chord-chord formula of angle, we get

$mHMK=\frac{1}{2}(mJL+mHLK)$

On putting the values in the above formula, we get

⇒ $(8x-1)=\frac{1}{2}(15x-36+3x+2)$

On applying cross-multiplication, we get

⇒ $2(8x-1)=18x-34$

⇒ $16x-2=18x-34$

On subtracting "18x" from both sides, we get

⇒ $16x-2-18x=18x-34-18x$

⇒ $-2x-2=-34$

On adding "2" both sides, we get

⇒ $-2x=-34+2$

⇒ $-2x=-32$

⇒ $x=\frac{32}{2}$

⇒ $x=16$

On putting the value of "x" in mHLK = (15x-36)°, we get

⇒ (15x-36)° = (15×16-36)°

⇒ = (240-36)°

⇒ = (204)°

So that mHLK = (204)°

4 0

2 years ago

James has a total of 66 dollars in his piggy bank. He only has one dollar bills and two dollar bills in his piggy bank. If there

gavmur [86]

Answer:

32 one-dollar bills.

Step-by-step explanation:

Let x represent one-dollar bills and y represent two-dollar bills.

He has a total of 49 bills. Therefore:

$x+y=49$

The total amount of money James has is 66. x is worth one dollar, while y is worth two dollars. Therefore:

$1x+2y=66\\x+2y=66$

We have a system of equations. Solve by substitution:

$x+2y=66\\x+y=49\\x=49-y\\(49-y)+2y=66\\y=17\\x=49-17=32$

Therefore, James has 32 one-dollar bills and 17 two-dollar bills.

Checking:

$32(1)+17(2)\stackrel{?}{=}66\\32(1)+17(2)\stackrel{\checkmark}{=}66\\\\32+17\stackrel{?}{=}49\\49\stackrel{\checkmark}{=}49$

5 0

2 years ago

Find the volume of a right circular cone that has a height of 5.9 in and a base with a diameter of 11.2 in. Round your answer to

schepotkina [342]

Answer:

Step-by-step explanation:

volume=1/3×πr²h

=1/3π×(11.2/2)²×5.9

=1/3×22/7×(5.6)²×5.9

≈1/3×22×5.6×8×5.9

≈1938.3 in³

6 0

2 years ago

Sarah kicked a ball in the air. The function

Aleksandr-060686 [28]

Answer: The ball hits the ground at 5 s

Step-by-step explanation:

The question seems incomplete and there is not enough data. However, we can work with the following function to understand this problem:

$f=30 t- 6t^{2}$ (1)

Where $f$ models the height of the ball in meters and $t$ the time.

Now, let's find the time $t$ when the ball Sara kicked hits the ground (this is when $f=0 m$ ):

$0=30 t- 6t^{2}$ (2)

Rearranging the equation:

$6t^{2}-30 t=0$ (3)

Dividing both sides of the equation by $6$ :

$t^{2}-5 t=0$ (4)

This quadratic equation can be written in the form $at^{2}+bt+c=0$ , and can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=1$

$b=-5$

$c=0$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(0)}}{2(1)}$ (6)

Solving we have the following result:

$t=5 s$ This means the ball hit the ground 5 seconds after it was kicked by Sara.

6 0

2 years ago

Consider the curve y=4x-x^3 and the chord AB joining points A(-3, 15) and B(2, -15) on the curve. Find the x- and y-coordinates

Elis [28]

We are asked to find the coordinates of the points where the slope of the curve is equal to the slope of the chord AB. We are given the coordinates of A and B. We can use these and the slope formula to find the slope of the chord.

Recall the slope formula is: $m= \frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }$
We label the points as follows (note that you could change the way you label the points as you will get the same slope regardless of what point is designated with the 1s and which is designated with the 2s).
$A=( x_{1} , y_{1)}=(-3,15)$
$B=( x_{2} , y_{2}=(2,-15)$
Plugging these into the slope formula yields: $m= \frac{-15-15}{2--3 }= \frac{-30}{5} =-6$
The slope of the chord is -6

Next consider the curve $y=4x- x^{3}$ . We can find the slope of the curve by taking the first derivative. Thus, the slope of the curve is given by: $y^{'} =4-3 x^{2}$

Let us set the slope of the curve equal to -6 to find the points (x,y) where the chord and the curve have the same slope.

That is, $-6=4-3 x^{2}$
$3 x^{2} =10$
$x^{2} = \sqrt{ \frac{10}{3} }$
$x= \sqrt{ \frac{10}{3} },x= -\sqrt{ \frac{10}{3} }$

We now have the x-coordinates of the points at which the slope of the chord equals that of the curve. To find the y-coordinates we substitute the values we found for x in the original equation as follows:

$y=4( \sqrt{ \frac{10}{3} }) -( \sqrt{ \frac{10}{3} }) ^{3} =4( \frac{10}{3}) ^{ \frac{1}{2} }-( \frac{10}{3}) ^{ \frac{3}{2} }$
and
$y=4( -\sqrt{ \frac{10}{3} }) -(- \sqrt{ \frac{10}{3} }) ^{3} =-4( \frac{10}{3}) ^{ \frac{1}{2} }+( \frac{10}{3}) ^{ \frac{3}{2} }$

Thus the two points we seek (x,y) are:
$( \sqrt{ \frac{10}{3} }, 4( \frac{10}{3}) ^{ \frac{1}{2} }-( \frac{10}{3}) ^{ \frac{3}{2} } )$
and
$( -\sqrt{ \frac{10}{3} }, -4( \frac{10}{3}) ^{ \frac{1}{2} }+( \frac{10}{3}) ^{ \frac{3}{2} } )$

4 0

2 years ago