(0,0) , (1,6) , (2,24) , (3,54) , (4,96) what is y ?

If v(t)=-16t+20, find v(1/2a+1)

Temka [501]

Possible derivation:

d/da(4 - 8 a)

Differentiate the sum term by term and factor out constants:

= d/da(4) - 8 (d/da(a))

The derivative of 4 is zero:

= -8 (d/da(a)) + 0

Simplify the expression:

= -8 (d/da(a))

The derivative of a is 1:

= -8 1

Simplify the expression:

Answer: = -8

4 0

3 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

WARRIOR [948]

Answer:

(a) The probability that more than 180 will take your free sample is 0.1056.

(b) The probability that fewer than 200 will take your free sample is 0.9997.

(c) The probability that a customer will take a free sample and buy the product is 0.2184.

(d) The probability that between 60 and 80 customers will take the free sample and buy the product is 0.8005.

Step-by-step explanation:

We are given that about 56% of all customers will take free samples. Furthermore, of those who take the free samples, about 39% will buy what they have sampled.

The day you were offering free samples, 303 customers passed by your counter.

Firstly, we will check that it is appropriate to use the normal approximation to the binomial, that is;

Is np > 5 and n(1-p) > 5

In our question, n = sample of customers = 303

p = probability that customers will take free sample = 56%

So, np = $303 \times 0.56$ = 169.68 > 5

n(1-p) = $303 \times (1-0.56)$ = 133.32 > 5

Since, both conditions are satisfied so it is appropriate to use the normal approximation to the binomial.

Now, mean of the normal distribution is given by;

Mean, $\mu$ = $n \times p$ = 169.68

Also, the standard deviation of the normal distribution is given by;

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{303 \times 0.56 \times (1-0.56)}$ = 8.64

Let X = Number of people who will take your free sample

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) The probability that more than 180 will take your free sample is given by = P(X > 180) = P(X > 180.5) {Using continuity correction}

P(X > 180.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{180.5-169.68}{8.64}$ ) = P(Z > 1.25) = 1 - P(Z < 1.25)

= 1 - 0.8944 = 0.1056

(b) The probability that fewer than 200 will take your free sample is given by = P(X < 200) = P(X < 199.5) {Using continuity correction}

P(X < 199.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{199.5-169.68}{8.64}$ ) = P(Z < 3.45) = 0.9997

(c) We are given in the question that of those who take the free samples, about 39% will buy what they have sampled, this means that we have;

P(Buy the product / taken a free sample) = 0.39

So, Probability(customer will take a free sample and buy the product) = P(customer take a free sample) $\times$ P(Buy the product / taken a free sample)

= 0.56 $\times$ 0.39 = 0.2184

(d) Now our mean and standard deviation will get changed because the probability of success now is p = 0.2184 but n is same as 303.

So, Mean, $\mu$ = $n \times p$ = $303 \times 0.2184$ = 66.18

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{303 \times 0.2184 \times (1-0.2184)}$ = 7.192

Now, the probability that between 60 and 80 customers will take the free sample and buy the product is given by = P(60 < X < 80) = P(59.5 < X < 80.5) {Using continuity correction}

P(59.5 < X < 80.5) = P(X < 80.5) - P(X $\leq$ 59.5)

P(X < 80.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{80.5-66.18}{7.192}$ ) = P(Z < 1.99) = 0.9767

P(X $\leq$ 59.5) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{59.5-66.18}{7.192}$ ) = P(Z $\leq$ -0.93) = 1 - P(Z < 0.93)

= 1 - 0.8238 = 0.1762

Therefore, P(59.5 < X < 80.5) = 0.9767- 0.1762 = 0.8005.

4 0

2 years ago

What is the answer for this problem

Margarita [4]

The answer is 2 and negative 9

3 0

2 years ago

Evaluate: (d + 1)(d - 1 + 6) d = 2

Allisa [31]

F(d)=(d+1)(d-1+6)
f(d)=(d+1)(d+5)
f(2)=(2+1)(2+5)
f(2)=(3)(7)
f(2)=21

3 0

2 years ago

A solid cuboid is made from centimeter cubes how many centimeter cubes were used to make the cuboid

prisoha [69]

This question is incomplete because it lacks the appropriate diagram.

Please find attached this answer the appropriate diagram.

Kindly note that: The diagram was obtained from or referenced from Planes and Elevation (F) Version 3 January 2016.

Answer:

72 centimeter cubes

Step-by-step explanation:

Looking at the attached diagram, we are given 3 views of the cuboid

a) The plan view: The is the view of the cuboid from the top or a very high position

b) Front Elevation view: This is the view of the cuboid from the front

c) Side elevation view: This is the view of the cuboid from the side.

In the referenced and attached diagram,

The plan view shows us the top of the cuboid with a length of 4 and breadth of 6

The front elevation view shows us the front of the cuboid with a length of 3 and breadth of 4

The side elevation view shows us the side of the cuboid with a length is 3 and breadth of 6

We are told to find the number of centimeter cubes were used to make the cuboid. This means we are to find the volume of the cuboid.

Volume of a cuboid = Length × Width × Height

Combining the 3 views together,

The Length of the cuboid = 4

Width of the cuboid = 6

Height of the cuboid = 3

Volume of the cuboid = 4 × 6 × 3

= 72

Therefore, the number of centimeter cubes that were used to make the cuboid is 72.

5 0

2 years ago