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2 years ago

He can you guys help me with this ?

Mathematics

1 answer:

antiseptic1488 [7]2 years ago

8 0

I’ll try my best but sorry if I can’t help a lot I’m not the best at math

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The length of the hypotenuse of a right triangle is 26 in. one leg measures 13 in. find the length of the other leg. round your

Ghella [55]

Use c^2=a^2+b^2
26^2= 13^2 + b^2
676=169 + b^2 (subtract 169 from both sides)
507= b^2 (take the square root of each side)
22.52 in. =b

3 0

2 years ago

What is the answer for 5x1+2?<br> I

tekilochka [14]

Answer:

Step-by-step explanation:

5 0

2 years ago

May I know this answer

kap26 [50]

I think it’s C but I may be wrong

3 0

2 years ago

The student body President of a high school claims to know the names of at least 1000 of the 1800 students at the school. To tes

hoa [83]

Answer:

The 99% confidence interval for p in this case is (0.3317, 0.5883).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Randomly selects 100 students from the school and asks the President to name each one. The President is able to correctly name 46 of the students.

This means that:

$n = 100, \pi = \frac{46}{100} = 0.46$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.3317$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 + 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.5883$

The 99% confidence interval for p in this case is (0.3317, 0.5883).

4 0

2 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

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He can you guys help me with this ?​

He can you guys help me with this ?