6. You roll a 6 sided die and flip a coin. What is the

TEA [102]

You want to start by subtracting 5 from both sides to get 2/3x=-8/3 then cross multiply to get whole numbers which is 6=-24x then your gonna want to isolate the variable so divide both sides by -24 to get 6/-24=x then you can simplify to x=-1/4

4 0

2 years ago

Is 5.78778777... A rational or irrational number

Minchanka [31]

It is irrational because it does not repeat a pattern nor does it stop.

8 0

3 years ago

SOLVE THE EQUATION FOR Y

max2010maxim [7]

$(4)\ -6x + y = 11\ \ \ \Rightarrow\ \ \ y=6x+11\\\\(5)\ \ 8x + 2y = 10\ \ \ \Rightarrow\ \ \ 2y=-8x+10\ \ \ \Rightarrow\ \ \ y=-4x+5\\\\(6)\ \ 6x - 3y = -9\ \ \ \Rightarrow\ \ \ -3y=-6x-9\ \ \ \Rightarrow\ \ \ y=2x+3\\\\(7)\ \ -4x + 2y = 16\ \ \ \Rightarrow\ \ \ 2y=4x+16\ \ \ \Rightarrow\ \ \ y=2x+8\\\\
(8)\ \ 10x - 5y = 25\ \ \ \Rightarrow\ \ \ -5y=-10x+25\ \ \ \Rightarrow\ \ \ y=2x-5\\\\(9)\ \ 3x + 2y = -8\ \ \ \Rightarrow\ \ \ 2y=-3x-8\ \ \ \Rightarrow\ \ \ y=-1.5x-4
$

6 0

3 years ago

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5-6}{0.2}$

$Z = -5$

$Z = -5$ has a pvalue of 0.

1 - 0 = 1

5 0

3 years ago

PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST

ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

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

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

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 = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

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

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

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

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

7 0

3 years ago