la raiz cuadrada de un numero negativo tiene solución en el campo numerico de los reales verdadero o falso

Complete the following steps for this equation.

Troyanec [42]

1. 4x – 20 = 5y

so 5y = 4x – 20

divide both sides by 5

y = (4/5)x - 4

2. x-intercept is at y=0

so (4/5)x - 4 = 0

x = 5

so it is (5,0)

3. y-intercept is at x=0, y = 0 - 4 = -4

so it is (0,-4)

6 0

3 years ago

PLEASE HELPPP MEEEEEEEEEE

qwelly [4]

Answer:

x = 17°

Step-by-step explanation:

Angles And Parallel Lines

If two parallel lines are crossed by a third line, corresponding angles are congruent. Also, consecutive angles are supplementary

We can clearly see in the image provided, being w and b parallel, that the corresponding angles (4x-3) and 65° are congruent. We can express that relation as

$4x-3=65$

Operating

$4x=65+3=68$

$x=\frac{68}{4}=17$

So the value of the angle x is 17°

8 0

3 years ago

7. The manager of the soda shop decides that because you are a good customer, you may use a flavor as many times as you’d like.

Black_prince [1.1K]

The different possible mixtures can you make is 1716 .

Step-by-step explanation:

Given that,

The total number of flavors available in the shop = 13 flavors
The number of flavors you need to choose = 6 flavors

Here, we have to use the formula for combination,

nCr = n! / r! × (n-r) !

where,

n is the total number of flavors
r is the number of flavors you need to choose

⇒ 13C6 = 13 ! / 6! × (13-6)!

⇒ 13! / (6! × 7! )

⇒ 13 × 12 × 11 × 10 × 9 × 8 × 7! / (6! × 7! )

⇒ 13 × 12 × 11 × 10 × 9 × 8 / 6!

⇒ 13 × 12 × 11 × 10 × 9 × 8 / 6 × 5 × 4 × 3 × 2 × 1

⇒ 1716

Therefore, you can 1716 different possible mixtures.

3 0

3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

$P = P_{1} + P_{2} + P_{3}$

$P_{1}$ is the probability that all three of them are 13-watt. So:

$P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277$

$P_{2}$ is the probability that all three of them are 18-watt. So:

$P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415$

$P_{3}$ is the probability that all three of them are 23-watt. So:

$P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173$

$P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865$

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

$P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245$

There is a 12.45% probability that one bulb of each type is selected.

3 0

3 years ago

Imagine we conduct a one-way independent ANOVA with four levels on our independent variable and obtain a significant result. Giv

fgiga [73]

Answer:

(C) Bonferroni

Explanation:

A Bonferroni test is a type of multiple comparison test used in statistical analysis. It is used when several dependent or independent statistical tests are being performed simultaneously.Bonferroni designed a test or an adjustment to prevent data from incorrectly appearing to be statistically significant.

It is used to counteract the problem of multiple Comparisons.

The Bonferroni correction is appropriate when a single false positive in a set of tests would be a problem. It is mainly useful when there are a fairly small number of multiple comparisons and you're looking for one or two that might be significant.

5 0

3 years ago