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r-ruslan [8.4K]
3 years ago
6

What is the domain and range of h(t)=cot t

Mathematics
1 answer:
NeTakaya3 years ago
7 0
Domain = R
Range = R .............
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Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to
polet [3.4K]

Answer:

(a) The probability that X is at most 30 is 0.9726.

(b) The probability that X is less than 30 is 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.

Step-by-step explanation:

We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.

Let X = <u><em>the number among these that are nonconforming and can be reworked</em></u>

The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).

Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.

Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).

So, the new mean of X, \mu = n \times p = 200 \times 0.11 = 22

and the new standard deviation of X, \sigma = \sqrt{n \times p \times (1-p)}

                                                                  = \sqrt{200 \times 0.11 \times (1-0.11)}

                                                                  = 4.42

So, X ~ Normal(\mu =22, \sigma^{2} = 4.42^{2})

(a) The probability that X is at most 30 is given by = P(X < 30.5)  {using continuity correction}

        P(X < 30.5) = P( \frac{X-\mu}{\sigma} < \frac{30.5-22}{4.42} ) = P(Z < 1.92) = <u>0.9726</u>

The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.

(b) The probability that X is less than 30 is given by = P(X \leq 29.5)    {using continuity correction}

        P(X \leq 29.5) = P( \frac{X-\mu}{\sigma} \leq \frac{29.5-22}{4.42} ) = P(Z \leq 1.70) = <u>0.9554</u>

The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is given by = P(15 \leq X \leq 25) = P(X < 25.5) - P(X \leq 14.5)   {using continuity correction}

       P(X < 25.5) = P( \frac{X-\mu}{\sigma} < \frac{25.5-22}{4.42} ) = P(Z < 0.79) = 0.7852

       P(X \leq 14.5) = P( \frac{X-\mu}{\sigma} \leq \frac{14.5-22}{4.42} ) = P(Z \leq -1.70) = 1 - P(Z < 1.70)

                                                          = 1 - 0.9554 = 0.0446

The above probability is calculated by looking at the value of x = 0.79 and x = 1.70 in the z table which has an area of 0.7852 and 0.9554.

Therefore, P(15 \leq X \leq 25) = 0.7852 - 0.0446 = 0.7406.

5 0
3 years ago
Please help, 50 points, will crown brainliest
VashaNatasha [74]
  • Answer:Q1 27.6363636          Q2 304/11 = q1 answer

Step-by-step explanation:

8 0
3 years ago
What is X in the equation -15x+15(-4)=-30
love history [14]

<em><u>Answer: x=-2 *The answer should be the negative sign.*</u></em>

Step-by-step explanation:

distributive property: a(b+c)=ab+ac

remove parenthesis

-15x-15*4=-30

15*4=60

-15x-60=-30

add 60 both sides of an equation.

-15x-60+60=-30+60

simplify.

-15x=30

divide by -15 both sides of an equation.

-15x/-15=30/-15

simplify.

30/15=2

x=-2

Hope this helps!

Thanks!

Have a great day!

7 0
3 years ago
Suppose y varies jointly as x and z. If y=15 when x=1/2 and z=6, what is the constant of variation?
Mashcka [7]

Answer:

y= 15

Step-by-step explanation:

Y= Kxz

y=6  x=1/2   z=6

15= K (1/2) (6)

15= K (3)

K= 5

SO

y= (5) (1/2) (6)

Y= 15

4 0
2 years ago
Read 2 more answers
Which values from the set{5,7,9,11,13} make the inequality w-4&lt;8 true?
Rasek [7]

Answer:

5, 7, 9, 11

Step-by-step explanation:

Solve the inequality

w - 4 < 8 ( add 4 to both sides )

w < 12

w ∈ { 5, 7, 9, 11 }

7 0
3 years ago
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