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maks197457 [2]
3 years ago
8

 How long must a ladder be to reach to top of a 20 foot wallif the ladder and the top wall form a 32 degree angle at the top. Ro

und to 2 decimal places .

Mathematics
1 answer:
seraphim [82]3 years ago
3 0
Check the picture  below.

make sure your calculator is in Degree mode.

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Suppose a fair six sided die has been relabeled so that the 6 is replaced with another 5,i.e., the sides read 1,2,3,4,5,5. If th
Anvisha [2.4K]

Answer:

P(rolling five distinct values) = 1/3888

Step-by-step explanation:

We are told that I the fair sided dice, 6 is replaced with 5.

Thus,the numbers are now;

1,2,3,4,5,5.

This means a total of 6 possible numbers appearing.

Now, in these 6 numbers there are only 5 distinct numbers since 5 occurs twice.

P(1) = 1/6

P(2) = 1/6

P(3) = 1/6

P(4) = 1/6

P(5) = 2/6

Thus,probability of rolling 5 distinct values = 1/6 × 1/6 × 1/6 × 1/6 × 2/6 = 1/3888

8 0
3 years ago
Solve for the value of w.
Bezzdna [24]
(4w-8)+(3w+6)=180
7w-2=180
7w=180+2
7w=182
w=26
7 0
2 years ago
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
How do you figure it out in steps 3r+n^2-r+5-2n+2
Katyanochek1 [597]
If you need to simplify the equation it would be:
n^{2}+2r-2n+7
If not sorry..
Hope This Helps!
8 0
4 years ago
Given:
Colt1911 [192]
Your question is very confusing but x=8 so just plug in 8 where you see X
3 0
3 years ago
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