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Anuta_ua [19.1K]
3 years ago
15

26=8+v como se resuelve?

Mathematics
1 answer:
posledela3 years ago
6 0
26 = 8 + v
<span>Restar -8 de cada lado
v = 18</span>
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The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding
crimeas [40]

Answer:

Step-by-step explanation:

Mean = (29.7 + 29.4 + 31.7 + 29.0 + 29.1 + 30.5 + 29.1 + 29.8)/8 = 29.7875

Mean = 29.7875 × 1000 = $29787.5

Standard deviation = √(summation(x - mean)²/n

n = 8

Summation(x - mean)² = (29.7 - 29.7875)^2 + (29.4 - 29.7875)^2 + (31.7 - 29.7875)^2 + (29.0 - 29.7875)^2 + (29.1 - 29.7875)^2 + (30.5 - 29.7875)^2 + (29.1 - 29.7875)^2 + (29.8 - 29.7875)^2 = 5.88875

Standard deviation = √(5.88875/8

s = 0.88

s = 0.88 × 1000 = $880

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0:µ ≤ 30000

For the alternative hypothesis,

H1:µ > 30000

This is a right tailed test.

b) The decision rule is to reject the null hypothesis if the significance level of 0.05 is greater than the probability value. If it is otherwise, we would fail to reject the null hypothesis.

c) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 29787.5

µ = population mean = 30000

s = samples standard deviation = 880

t = (29787.5 - 30000)/(880/√8) = - 0.68

We would determine the p value using the t test calculator. It becomes

p = 0.26

d) Since alpha, 0.1 < the p value, 0.26, then we would fail to reject the null hypothesis.

6 0
3 years ago
Prove that 3(x+1)(x+7)-2x+5)^2 is never positive
lisov135 [29]

Answer:

Yes, 3(x+1)(x+7)-(2x+5)² is never positive

Step-by-step explanation:

So....

3(x+1)(x+7)-(2x+5)(2x+5)

=3(x²+8x+7)-(4x²+20x+25)

=3x²+24x+21-4x²-20x-25

=-x²+4x-4

5 0
2 years ago
If x+y=12 and xy=15,find the value of (x^2+y^2)
Mice21 [21]

Answer:  x^2+y^2=\dfrac{105\pm 18\sqrt6}{2}

<u>Step-by-step explanation:</u>

EQ1:  x + y = 12     --> x = 12 - y

EQ2:  xy = 15      

Substitute x = 12-y into EQ2 to solve for y:

(12 - y)y = 15

12y - y² = 15

0 = y² - 12y + 15

    ↓     ↓         ↓

  a=1   b= -12   c=15

.\  y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\quad =\dfrac{-(-12)\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}\\\\\\.\quad =\dfrac{12\pm \sqrt{144-120}}{2}\\\\\\.\quad =\dfrac{12\pm \sqrt{24}}{2}\\\\\\.\quad =\dfrac{12\pm 2\sqrt{6}}{2}\\\\\\.\quad =6\pm \sqrt{6}

Now, let's solve for x:

xy=15\\\\x(6\pm\sqrt6)=15\\\\x=\dfrac{15}{6\pm\sqrt6}\\\\\\x=\dfrac{15}{6\pm\sqrt6}\bigg(\dfrac{6\pm\sqrt6}{6\pm\sqrt6}\bigg)=\dfrac{6\pm \sqrt6}{2}

Lastly, find x² + y² :

y^2=(6\pm \sqrt6)^2\quad \rightarrow \quad y^2=36\pm 12\sqrt6 +6\quad \rightarrow \quad y^2=42\pm 12\sqrt6

x^2=\bigg(\dfrac{6\pm \sqrt6}{2}\bigg)^2\quad \rightarrow \quad x^2=\dfrac{42\pm 12\sqrt6}{4}\quad \rightarrow \quad x^2=\dfrac{21\pm 6\sqrt6}{2}

                                                                                 

x^2+y^2=\dfrac{21\pm 6\sqrt6}{2}+42\pm 12\sqrt6\\\\\\.\qquad \quad = \dfrac{21\pm 6\sqrt6}{2}+\dfrac{84\pm 24\sqrt6}{2}\\\\\\. \qquad \quad = \dfrac{105\pm 18\sqrt6}{2}

5 0
3 years ago
Write down a number that is not a rational number. explain why.
KatRina [158]
3.7491739193721... because rational numbers are numbers like -3 -2 -1 0 1 2 3 and decimals like 3.3333333 and 1.62162162162162 because they repeat the same numbers like a pattern, with 3 repeating or 162.The decimal 3.7491739193721... is not rational, or irrational because it is not repeating or terminating, or stopped.
8 0
2 years ago
The cost of watching videos on a website is represented by C(a) = 3.00a + 4.00, where a is the number of videos watched. The cos
aksik [14]

Answer:

The answer is A; H(a) = 3.00a +19.00; shift 15 units up

Step-by-step explanation:

If the downloading feature is enabled the cost function becomes H(a) = 3.00a + 19.00

The y-interpret of H(a) is 15 greater than the y-intercept if C(a)

The slopes of both functions are the same

3 0
3 years ago
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