Does anyone know how to do this ?

Circle O and triangle OQR are shown below. What is the length of OQ?

zysi [14]

Answer:

Step-by-step explanation:

A circle is inscribed in an equilateral triangle PQR with centre O. If angle OQR = 30°, what is the perimeter of the triangle?

This is a circle inscribed in an equilateral triangle with side s.

If you are asking for the perimeter of PQR, it is 3s.

If you are asking for the perimeter of OQR, it is (3+23–√3)s

Since OR and SR are the hypotenuses of right triangles with adjacent side equal to ½ s, their length is ½s / cos 30° = (√3) /3.

(3/3)s + ((√3) /3)s + ((√3) /3)s = ((3 + 2√3)/3)s ≈ 2.1547s

Hope it helps

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3 0

2 years ago

A study of the number of business lunches that executives in the insurance and banking industries claim as deductible expenses p

gizmo_the_mogwai [7]

Answer:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

$p_v =2*P(t_{(88)}>2.604)=0.0108$

So the p value is a very low value and using any significance level for given $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

$\bar X_{1}=9.1$ represent the mean for the sample 1

$\bar X_{2}=8$ represent the mean for the sample 2

$s_{1}=1.9$ represent the sample standard deviation for the sample 1

$s_{2}=2.1$ represent the sample standard deviation for the sample 2

$n_{1}=40$ sample size for the group 1

$n_{2}=50$ sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

Statistical decision

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=40+50-2=88$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(88)}>2.604)=0.0108$

So the p value is a very low value and using any significance level for given $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

3 0

3 years ago

A bag contains 6 red balls and blue balls. 4 balls are selected at random. find the probability of selecting 4 red balls.

masya89 [10]

Another way is to note that there are (104)(104) (“10 choose 4”) ways to select 4 balls from a collection of 10. If 4 of those 10 balls are “special” in some way (in this case, “special” = “red”), then the number of ways to choose 4 special balls is (44)(44). (The factor of (60)(60) is included to convey that, after choosing 4 special balls, we choose none of the 6 non-special balls.) This line of reasoning gives the second expression.

8 0

3 years ago

Sam has $440 in $20, $10, and $5 dollar bills. The number of twenties, tens and fives are consecutive integers in this order. Ho

Ludmilka [50]

Answer: There are 12 bills for $20, 13 bills for $10 and 14 bills for $5.

Step-by-step explanation:

Let the number of twenties be x

Let the number of tens be x+1

Let the number of fives be x+2

Amount from twenties will be given as

$20\times x=20x$