Use 3 darts. Each dart must hit the board.

Consider the probability that at least 93 out of 154 CDs will not be defective. Assume the probability that a given CD will not

Yuki888 [10]

Answer:

0.5962

Step-by-step explanation:

Given that :

p = 61% = 0.61

q = 1 - p = 1 - 0.61 = 0.39

n = 154 ; x = 93

Using the binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

P(x>=93) = p(x=93)+p(x=94)+...+p(x=n)

P(x>= 93) = 0.59619

P(x>= 93) = 0.5962

7 0

3 years ago

Please answer quickly!!!

yKpoI14uk [10]

Answer:

$c=20$

Step-by-step explanation:

The diagonals of a rectangle always bisect each other. Therefore, we can set $QT=ST$ :

$c=4c-60,\\-3c=-60,\\c=\fbox{20}$

3 0

3 years ago

Tell whether the pair of polygons is similar. Explain why or why not. (3 points) I get the general idea of this but I’m having t

masha68 [24]

Answer:

PQRS and TUVW are similar.

Step-by-step explanation:

We are give two polygons.

Sides of first polygon are :

PQ = 8 ft

QR= 13ft

RS = 8ft

PS = 13ft.

Sides of second polygon:

TW = 2.4 ft

TU = 5.2 ft

UV = 2.4 ft

VW = 5.2 ft.

We can see that PQRS and TUVW are parallelograms.

Note: If an angle of a parallelogram is equal to an angle of another parallelogram then parallelograms would be similar.

We can see that <Q ≅ <T = 120°.

<h3>Therefore, PQRS and TUVW are similar.</h3>

5 0

3 years ago

Read 2 more answers

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

Dafna1 [17]

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

L'(t) = 34.416*e^(-0.18*t)

L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

6 cm/yr = 34.416*e^(-0.18*t)

e^(0.18*t) = 34.416 / 6

0.18*t = Ln(34.416/6)

t = Ln(34.416/6) / 0.18

t = 10 year

7 0

3 years ago

The product of 8/15, 6/5,1/3 is

Debora [2.8K]

Product is the result of multiplying the numbers, so the product of 8/15, 6/5,1/3 means that we have to multiply the three numbers, like this:

$\frac{8}{15} * \frac{6}{5} * \frac{1}{3}$
let's make them into one fraction:

$\frac{8*6*1}{15*5*3}$

We can simplyfy:

 $\frac{8*6}{15*5*3}$


 $\frac{8*2}{15*5}$

and multiply:

 $\frac{16}{75}$

and this is the correct answer!

3 0

4 years ago