4. What is the next number? 0 3 8 15 24 35 .......

There are 4 jacks and 13 clubs in a standard, 52-card deck of playing cards. What is the probability that a card picked at rando

ale4655 [162]

Answer:

16/52, or 4/13.

Step-by-step explanation:

First, since we know that the question is asking for the probability of a club or a jack, we know that we have to add the two probabilities. The first probability is that of picking a club, which is 13/52. The probability of picking a jack (be sure not to overlap; don't double count the jack of clubs) is 3/52. Adding these two gives us 13/52+3/52=16/52, which simplifies to 4/13.

3 0

4 years ago

2. Find the solution set of the inequality:

m_a_m_a [10]

Answer:

The answer is B

Step-by-step explanation:

I got a giid grade

8 0

3 years ago

1.Dakota bought 1.7 pounds of ham, 3.15 pounds of onions, and 6 cans of soup. What was the total cost before sales tax? Round

MrRa [10]

Answer: 15.73

Step-by-step explanation:

4 0

3 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for n = k ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for n = k + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

6 0

2 years ago

Read 2 more answers

When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w

deff fn [24]

Answer:

A 90% confidence interval for p will be narrower than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - α) % confidence interval for a population proportion is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Here $\hat p$ is the sample proportion.

The margin of error of the confidence interval is:

$MOE= z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The MOE is dependent on:

Confidence level
Standard deviation
Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of z for 90% confidence level is:

$z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645$

And the critical value of z for 99% confidence level is:

$z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96$

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for p will be narrower than the 99% confidence interval.

6 0

3 years ago