Mathematics make up exam questions Answers solved.

Solution: (1) There are 17 female students, 5 of whom walk to school every day, indicating the necessary conditions: 17-5=12;

7 people have bikes available, but not necessarily; 9 people can travel by car, but not necessarily by car; Herein lies the crux of the matter. It shows that there are bicycles and can ride: 7 + 9-12 = 4 people with two conditions.

Therefore, sometimes there are 4 people on a bicycle and sometimes by car.

2) The probability of both positive and negative occurring for the fifth time is 50%; There are only two situations in which a coin toss occurs: heads or tails, and the probability before there is no toss is 1 2=50%.

T1 is because people who have bicycles sometimes ride and sometimes ride in cars.

Sometimes the people who ride and sometimes ride are:

4 people. T2 tosses a coin, each time there is a chance that 1 2 is heads or tails, four times in a row, that is.

Positive --- negative.

Positive and negative Positive and negative.

Positive and negative positive and negative positive and negative.

positive and negative positive and negative positive and negative positive and negative positive and negative.

So the probability of exactly two heads is 5 16, and the probability of at least one heads is 15 16, count and count you.

Let's talk about the second question first, no matter how many times you appear in front of you, the probability of both tails and heads is 50%, so what they said is not right, and I will add to the first question later.

Supplement: 5 + 7 + 9 = 21 21-17 = 4 people Answer: Sometimes there are 4 people who ride a car and sometimes a bicycle.

Reason: It is true that 5+7+9 is greater than 17, but the extra people are those who sometimes ride cars and sometimes ride bicycles.

I hope I don't want your share either.

5 of them walk to school every day, and 12 girls travel to school by bicycle or car.

5+7+9 is greater than 17 because some people sometimes ride bicycles and sometimes take cars to school.

5+7+9-17=4, then there are 4 people who sometimes ride bicycles and sometimes take cars to school.

2. The probability of the 5th occurrence of heads or tails is 50%, because the 5th and the previous 4 events are independent of each other, and the occurrence of the first 4 events does not affect the outcome of the 5th event.

1. The reason is that there is an intersection between people who have bicycles and people who can take cars, that is, some of them sometimes ride bicycles and sometimes take cars. Such a number of people is 7 + 9 - (17 - 5) = 4 people.

2. The probability of the 5th occurrence of heads and tails is 50%. Probabilities do not change as an event occurs.

1 3, Mention x If you don't understand it yet, the molecular part is approximate to x when x tends to 0.

I don't understand the first question.

First sold 300 grams, and then said that 15 boxes were sold, and it is said that each box should be 300 15 = 20 grams = kilograms, but is there such a small box?Check to see if you have made a mistake in the question.

The second assumes that the Earth's equator is x 10,000 kilometers.

30=x*8-2

8x=32x=4 supplementary questions.

Multiply both sides by 3

x-5=x=

Help you do it all?

This is the first year of mathematics.

You can send me an electronic copy and I'll do it for you.

Take it a ride.

<> such as the oak base mu Liang Sen Tu Feng judgment.

Didn't save the big brother, don't struggle.

Solution: definite integral: definite integral of x 2:

1 3 x 3,+c e x=e x+c sinx points=-cosx+c, 1 2x 3 2-1 3x 3+c, 1 2x 2+x+c, I can't see the following.

The problem of bump intervals and inflection points is relatively simple, first of all, the problem of derivation, and secondly, the positive and negative relationship between bumpiness and the second derivative function.

This is very easy, after the derivation, 0 is the concave point, and after the derivation<0 is the bump.

The standing point is derivative=0, and the inflection point is after two derivatives=0

Just take a look at the formula in the book again.

It is sufficient to determine the sign of the second derivative.