Answer:
This is how tackle this type of problem:
Draw a right triangle with the indicated sides: side opposite the angle = 12 and hypotenuse = 13
Then, calculate the 3rd side by Pythagoras: x2 + 122 = 132
x2 = 169 - 144
x2 = 25
x = 5
Now, we have all 3 sides of our diagram triangle, so we can find all the trig ratios for the angle.
Next, we use the double angle formula to simplify cos(x + 30) = cos x cos 30 - sin x sin 30
= 5/13 . √3/2 - 12/13 . 1/2
= 5√3 /26 - 6/13
= (5√3 - 12) / 26
Step-by-step explanation:
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Answer:
42.08inches
Step-by-step explanation:
Given data
SIze of TV= 50 inch
Width= 27inches
Length= ??
Applying Pythagoras theorem
z^2= x^2+y^2
50^2= x^2+27^2
square root both sides
50^2= x^2+27^2
2500= x^2+ 729
2500-729=x^2
1771=x^2
x= √1771
x= 42.08
Hence the lenght is 42.08inches
Answer: x= -19/14 —> (19 over 14)
Answer:
The standard deviation of the sample mean is 4 minutes
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
The standard deviation of the population is 32.
Sample of 64.
So

The standard deviation of the sample mean is 4 minutes
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have
,
,
and
,
,
. The pooled estimate is given by
a. We want to test
vs
(two-tailed alternative).
The test statistic is
and the observed value is
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)