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2 years ago

If A is a prime factor of 36, and B is a prime factor of 15, what is the greatest possible difference between A and B?

Mathematics

1 answer:

Lelechka [254]2 years ago

4 0

Answer:

Step-by-step explanation:

Prime factorization

36 = 3 * 3 *2 * 2

15 = 3 * 5 The biggest difference is 5 -2 = 3

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PLEASE HELP ME PLZ!!

vichka [17]

Answer:

1. Formula is A2 : A9 = COUNT( A2: A9 ) = 8

2. Formula is SUM( A2: A9 ) = 36

3. Formula is B2 : B9 = COUNT( B2: B9) = 8

4. Formula is MAX( C2: C9) = 5

5. Formula is MIN( C4: C8) = 3

6. Formula is SUM( C5 - C6) = 0

7. Formula is AVERAGE( C2: C9) = 4

Step-by-step explanation: Have a nice day! ✌️

7 0

2 years ago

A standard roulette wheel with slots labeled 0, 00, 1, 2, 3, … , 36 is spun 60 times. Of these spins, the number 7 is spun 7 tim

WARRIOR [948]

Since the p-value of the test is of 0.00001 < 0.01, these results have statistical significance.

<h3>What is the relation between the p-value and the conclusion of the test hypothesis?</h3>

Depends on if the p-value is less or more than the significance level:

If it is more, the null hypothesis is not rejected, which means that the results do not have statistical significance.

If it is less, it is rejected, which means that the results have statistical significance.

In this problem, the probability is the p-value, hence since the p-value of the test is of 0.00001 < 0.01, these results have statistical significance.

More can be learned about p-values at brainly.com/question/13873630

#SPJ1

6 0

2 years ago

Consider the following information related to a set of quantitative sample data:

svlad2 [7]

Answer:

(a) There are outliers

(b) $x$ and $x >62$

Step-by-step explanation:

Given

$\sigma = 14.92$

$\bar x = 22.0$

$q_0 = -24$

$q_1 = 14.5$

$q_2 = 24.5$

$q_3 = 33.5$

$q_4 = 64$

Solving (a): Check for outliers

This is calculated using:

$Lower = Q_1 - (1.5 * IQR)$ --- lower bound of outlier

$Upper = Q_3 +(1.5 * IQR)$ --- upper bound of outlier

Where

$IQR = Q_3 - Q_1$

So, we have:

$IQR = 33.5 - 14.5$

$IQR = 19$

The lower bound of outlier becomes

$Lower = Q_1 - (1.5 * IQR)$

$Lower = 14.5 - (1.5 * 19)$

$Lower = 14.5 - 28.5$

$Lower = -14$

The upper bound of outlier becomes

$Upper = Q_3 +(1.5 * IQR)$

$Upper = 33.5 + 1.5 * 19$

$Upper = 33.5 + 28.5$

$Upper = 62$

So, we have:

$-14 \le x \le 62$ --- the range without outlier

Given that:

$q_0 = -24$ --- This represents the lowest data

$q_4 = 64$ --- This represents the highest data

-24 and 64 are out of range of $-14 \le x \le 62$ .

Hence, there are outliers

Solving (b): The outliers

The outliers are data less than the lower bound (i.e. less than -14) or greater than the upper bound (i.e. 62)

So, the outliers are:

$x$ and $x >62$

4 0

3 years ago

Given: cosθ= -4/5 , sin x = -12/13 , θ is in the third quadrant, and x is in the fourth quadrant; evaluate tan 1/2 θ

mrs_skeptik [129]

Answer:

Step-by-step explanation:

If you're looking for what the half angle of the tangent of theta is, I'm a bit confused as to why you think the angle in the 4th quadrant, x, is relevant. But maybe you don't know it isn't and it's a "trick" to throw you off. Hmm...

Anyways, the half angle identity for tangent is

$tan(\frac{\theta}{2})=\frac{sin\theta}{1+cos\theta}$

There are actually 3 identities for the tangent of a half angle, but this one works just as well as either of the others do, so I'm going with this one.

If theta is in QIII, the value of -4 goes along the x axis and the hypotenuse is 5. That makes the missing side, by Pythagorean's Theorem, -3. Filling in our formula:

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{1+(-\frac{4}{5}) }$ which simplifies a bit to