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3 years ago

On which attempts was his time less than 60 seconds?

Mathematics

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

4 6 and 7?

Step-by-step explanation:

all you have to do is find 60 sec mark and all attempts bellow it are below 60 sec or under 60 sec

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A job fair was held at the Student Union. 25% of the students who attended received job offers. Of all of the students at the jo

vovangra [49]

Answer:

A) Both events are not independent.

B) Both events are not mutually exclusive

C) 8.33%

D) 80%

Step-by-step explanation:

A) Both events are not independent. This is because, If B occurs it means that it is very likely that J will occur as well.

B) Both events are not mutually exclusive. This is because it is possible for both events J and B to occur at the same time.

C) we want to find the probability that Joe who is not a business student will receive the job offer.

This is;

P(J|Not B) = P(J & Not B)/P(Not B)

Now,

P(J & Not B) = P(J) – (P(B) × P(J | B))

25% of the students who attended received job offers. Thus; P(J) = 0.25

40% were from the College of Business. Thus;

P(B) = 0.4

Among the business students, 50% received job offers. Thus;

P(J|B) = 0.5

Thus;

P(J & Not B) = 0.25 - (0.4 × 0.5)

P(J & Not B) = 0.25 - 0.2

P(J & Not B) = 0.05

Since P(B) = 0.4

Then, P(Not B) = 1 - 0.4 = 0.6

Thus;

P(J|Not B) = 0.05/0.6

P(J|Not B) = 0.0833 = 8.33%

D) This probability is represented by;

P(B | J) = P(B & J)/P(J)

P(B & J) = (P(B) × P(J | B)) = (0.4 × 0.5) = 0.2

P(B | J) = 0.2/0.25

P(B | J) = 0.8 = 80%

5 0

3 years ago

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

3 years ago

Write the percent as a fraction or mixed number in simplest form.

ddd [48]

As a fraction the answer would be 5/16

8 0

3 years ago

Read 2 more answers

Please help with angle x

vodomira [7]

Answer:

x = 155

Step-by-step explanation:

4 0

2 years ago

So far you have completed 80% of the problems for your math exam. Your exam

zmey [24]

Answer:

Problems left to complete = 9

Step-by-step explanation:

Given that:

Number of problems on the test = 45

Percent of completed problems = 80%

Number of completed problems = $\frac{80}{100}*45$

Number of completed problems = 0.8 * 45 = 36

Problems left to complete = 45 - 36

Problems left = 9

Hence,

Problems left to complete = 9

5 0

3 years ago

Read 2 more answers