What is the length of the hypoyenuse of a right triangle if each of the two legs is 4 units?

Find the volume of a pyramid with a square base, where the side length of the base is

Orlov [11]

Answer:

2896.34 cubi centimetres

Step-by-step explanation:

7 0

3 years ago

What is the likelihood that a fair coin will land heads or tails?

Marina CMI [18]

Answer:

I believe it is 0.5

Step-by-step explanation:

If you flip a normal coin (called a “fair” coin in probability parlance), you normally have no way to predict whether it will come up heads or tails. Both outcomes are equally likely. There is one bit of uncertainty; the probability of a head, written p(h), is 0.5 and the probability of a tail (p(t)) is 0.5. The sum of the probabilities of all the possible outcomes adds up to 1.0, the number of bits of uncertainty we had about the outcome before the flip. Since exactly one of the four outcomes has to happen, the sum of the probabilities for the four possibilities has to be 1.0. To relate this to information theory, this is like saying there is one bit of uncertainty about which of the four outcomes will happen before each pair of coin flips. And since each combination is equally likely, the probability of each outcome is 1/4 = 0.25. Assuming the coin is fair (has the same probability of heads and tails), the chance of guessing correctly is 50%, so you'd expect half the guesses to be correct and half to be wrong. So, if we ask the subject to guess heads or tails for each of 100 coin flips, we'd expect about 50 of the guesses to be correct. Suppose a new subject walks into the lab and manages to guess heads or tails correctly for 60 out of 100 tosses. Evidence of precognition, or perhaps the subject's possessing a telekinetic power which causes the coin to land with the guessed face up? Well,…no. In all likelihood, we've observed nothing more than good luck. The probability of 60 correct guesses out of 100 is about 2.8%, which means that if we do a large number of experiments flipping 100 coins, about every 35 experiments we can expect a score of 60 or better, purely due to chance.

6 0

3 years ago

Read 2 more answers

Solve the equation 13-2x=-7

patriot [66]

Subtract 13 from both sides.

-13
-7 = -20

-2x = -20

Divide by 2 on both sides.
Remember: Negative divided by a negative makes a positive.

-2/-2 = 1
-20/-2 = 10

Your answer:
x = 10

8 0

3 years ago

Read 2 more answers

A=5/6(B+C) SOLVE FOR C

Shalnov [3]

A = 5/6(B + C)

multiply 6 on both sides

6A = 5(B + C)

divide 5 from both sides

6A/5 = B + C

subtract B from both sides

6A/5 - B = C

3 0

3 years ago

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated

cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

Let $p_1$ = proportion of cockroaches that died on glass surface.

$p_2$ = proportion of cockroaches that died on plasterboard surface.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of cockroaches that died on glass surface = $\frac{18}{36}$ = 0.50

$\hat p_2$ = sample proportion of cockroaches that died on plasterboard surface = $\frac{25}{36}$ = 0.694

$n_1$ = sample of cockroaches on glass surface = 36

$n_2$ = sample of cockroaches on plasterboard surface = 36

So, test statistics = $\frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36} } }$

= -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0

3 years ago