2/3 • 3/2 (3x+6)=4x+3•3/2

How many baseballs would fit into a box that has the dimensions (3*5)*2

Snezhnost [94]

3*5= 15

15*2= 30

30 baseballs

5 0

3 years ago

Please help..I dont get this and please Dont put a Link below please.

lianna [129]

Answer:

x = 27, y = 74

Step-by-step explanation:

Hi again :)

Since the angles of 5x + 4 and the one adjacent to x = 14 are corresponding angles, they are equal. That means that (5x + 4) + (x + 14) = 180 degrees.

5x + 4 + x + 14 = 180

6x + 18 = 180

6x = 162

x = 27

Also, the angle (5x + 4) and (2y - 9) are vertical angles, they are equal in value. We know the value of x now, so we'll substitute that in and solve for y.

5(27) + 4 = 2y - 9

135 + 4 = 2y - 9

139 = 2y - 9

148 = 2y

y = 74

As always, lmk if you have questions.

3 0

3 years ago

A number decreased by 6 gives twice the number

alexgriva [62]

Answer:

I think it's 4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Can u help solve this

8_murik_8 [283]

Answer and Step-by-step explanation:

The expression is equal to 90.

This is because the lines that cut the right angle are cutting it in half, so those base angles are 45, leaving the center angle to be 90 (90 + 45 + 45 = 180).

90 = 19x + 4.5

85.5 = 19x

x = 4.5

The value of x that makes the parallelogram a square is 4.5.

I hope this helps!

#teamtrees #PAW (Plant And Water)

6 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

(c) $P(X > 980) = 0.00001\\\\$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

3 years ago