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3 years ago

15

Could somebody help me with this please

2 answers:

alekssr [168]3 years ago

7 0

Answer:

x = 19°

Step-by-step explanation:

.........................

Forgive me if wrong

Arlecino [84]3 years ago

3 0

X is 19 decrees this is simple!

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Please guys help please

almond37 [142]

Wild guess a I’m only guessing

4 0

3 years ago

Appreciate answers and any help!

Savatey [412]

To find the expected value of the distribution, we multiply each outcome by it's probability. Doing this, we get that the expected value of defects on a skateboard is of $\frac{4}{25}$ .

Outcomes and probabilities:

0 defects, 9/10 probability

1 defect, 1/20 probability

2 defects, 1/25 probability

3 defects, 1/100 probability.

Expected value:

$E(X) = 0\frac{9}{10} + \frac{1}{20} + 2\frac{1}{25} + 3\frac{1}{100} = \frac{1}{20} + \frac{2}{25} + \frac{3}{100} = \frac{5 + 8 + 3}{100} = \frac{16}{100}$

Dividing both numerator and denominator by 4:

$\frac{4}{25}$

Thus, the expected value of defects on a skateboard is of $\frac{4}{25}$ .

A similar problem is given at: brainly.com/question/23156292.

4 0

3 years ago

If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after t seconds is gi

Contact [7]

Step-by-step explanation:

If a cannonball is shot directly upward with a velocity of 160 ft per second. Its height as a function of time is given by :

$h(t)=160t-16t^2$ .......(1)

t is in seconds

(a) Velocity is given by :

$v=\dfrac{dh}{dt}\\\\v=\dfrac{d(160t-16t^2)}{dt}\\\\v=(160-32t)\ ft/s$

Acceleration is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(160-32t)}{dt}\\\\a=-32\ ft/s^2$

(b) For maximum height put $\dfrac{dh}{dt}=0$

i.e.

$160-32t=0\\\\t=5\ s$

Put t = 5 s in equation (1). So,

$h(5)=160t-16t^2\\\\h(5)=160(5)-16(5)^2\\\\h(5)=400\ ft$

(c) When the ball reaches ground, its height is equal to 0. So,

h(t) = 0

$160t-16t^2=0\\\\t(160-16t)=0\\\\t=0,t=10\ s$

Hence, this is the required solution.

4 0

3 years ago

The vertices of ABC are A(-2, 2), B(6,2), and CO, 8). The perimeter of ABC?

Alexxx [7]

Answer:

Perimeter = 22.809836694575

Step-by-step explanation:

A(-2, 2) ; B(6,2) ; C(0, 8)

$AB=\sqrt{\left( 6--2\right)^{2} +\left(2-2 \right)^{2} } =\sqrt{64} =8$

$AC=\sqrt{\left( 0--2\right)^{2} +\left(8-2 \right)^{2} } =\sqrt{40} =2\sqrt{10}$

$BC=\sqrt{\left( 0-6\right)^{2} +\left(8-2 \right)^{2} } =\sqrt{72} =6\sqrt{2}$

Then

The perimeter of ΔABC = AB + BC + AC

$=8+6\sqrt{2} +2\sqrt{10}$

= 22.809836694575

3 0

3 years ago

Identify the coefficient in each term

Archy [21]

Answer:

5. 4

6. 1

7. 2

8. -1

Step-by-step explanation:

Coefficient is usually the whole number in the beginning. I believe 5-7 are correct. I'm a little iffy about 8. Good luck!

8 0

3 years ago