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sammy [17]
3 years ago
13

the average of two numbers is 34 the first number is three times the second number what are the two numbers

Mathematics
1 answer:
tino4ka555 [31]3 years ago
7 0
Hi there!

To solve this problem, we need to set up two equations and use the system of equations to solve.

Let x be the first number.
Let y be the second number.

(x + y) ÷ 2 = 34
x = 3y

Now, we can use substitution to solve.

(3y + y) ÷ 2 = 34
4y ÷ 2 = 34
2y = 34
y = 17

Now, we plug the value of y in the equation to solve for x.

x = 3*17
x = 51

Hope this helps!
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Step-by-step explanation:

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There are 3000 people at a concert you survey a random sample of 200 people and find that for 35 of them this is their first con
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Ok so
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What values for q (0 ≤q≤2π)<br> satisfy the equation?<br><br> 22√sin q + 2 = 0
Vesna [10]
Answer:
\frac{3 \pi }{4} , \frac{7 \pi }{4}

Explanation:
2√2 sin(q) + 2 = 0
2√2 sin(q) = -2
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sin(q) = \frac{- \sqrt{2} }{2}

Now, we know that:
sin (45) = \frac{ \sqrt{2} }{2}

From the ASTC rule, we know that the sine function is negative in the third and fourth quadrant.
This means that:
either q = 90 + 45 = 135° which is equivalent to \frac{3 \pi }{4}
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Hope this helps :)
6 0
3 years ago
Read 2 more answers
To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t
nikklg [1K]

Answer:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_{1}=19.1 represent the mean for the sample mean for 1

\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

Null hypothesis:\mu_{1} \geq \mu_{2}

Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 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 info given like this:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

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

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

Conclusion

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

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