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Basile [38]
2 years ago
8

Which expression is equivalent to square root of -80

Mathematics
2 answers:
Yuki888 [10]2 years ago
7 0

Answer:

The equivalent expression is (4\sqrt{5})i

Step-by-step explanation:

Given the expression \sqrt{-80} we can use prime factorization and a property of imaginary numbers to obtain an equivalent expression.

Let's start applying prime factorization over 80

80=(2).(40)=(2).(2).(20)=(2).(2).(2).(10)=(2).(2).(2).(2).(5)=(2^{4}).(5)

The other property is i^{2}=-1 (This is a property of imaginary numbers).

We can write -80 as

-80=(2^{4}).(5).i^{2}

Now we apply square root to the expression ⇒

\sqrt{(2^{4}).(5).i^{2}}=(4\sqrt{5})i

Law Incorporation [45]2 years ago
5 0
Sqrt(-80) = 8.94427191i
Hope it helped :d
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Step-by-step explanation:

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How many numbers from 10 to 99 have a tens place exactly 3 times greater than their ones place? PLZZZ answer this question . I w
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Answer:

45

Step-by-step explanation:

2 digit number starts from 10 ends at 99

between 10 and 19 there is only one number whose tens digit is more than ones digit.

that is 10

between 20 and 29 there are two numbers

20 and 21

like the same

between 30 and 39 there are 3 numbers

10–19. 1

20–29. 2

30–39. 3

40–49. 4

50–59. 5

60–69. 6

70–79. 7

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sum of first n natural numbers is n(n+1)/2

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Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
Harlamova29_29 [7]

Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0
2 years ago
Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou
monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

              B) 0.6796

              C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )

                   = P ( Z > 2.33 ) = 0.0099

                   = 0.9%

B) percentage of oranges weighing less than 8.7 ounces

  P( x < 8.7 ) = P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )

                    = P ( Z < 0.4667 ) = 0.6796

                    = 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = P (\frac{6.2-8}{1.5} <  \frac{x - u}{ std} < \frac{7-8}{1.5} )

                          = P ( -1.2 < Z < -0.66 )

                          = Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0
2 years ago
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