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Fed [463]
2 years ago
6

Joseph is creating a garden up against his fence in his backyard in the shape of a semi-circle. The diameter of his garden is 8

feet long. How much square footage of soil will he need to buy?
Plz help me I cant figure it out T_T
Mathematics
1 answer:
Marizza181 [45]2 years ago
7 0

Answer:

25 inches of soil

Step-by-step explanation:

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I really need help with this sampling question and it would mean alot if someone could answer it
GuDViN [60]

Answer:

vc vzc vzv czzzzzzzzzzzz

Step-by-step explanation:

vczzzzzzzzzzz            vdfBbbbbbbbbbfbfc

7 0
3 years ago
Suppose a production line operates with a mean filling weight of 16 ounces per container. Since over- or under-filling can be da
miss Akunina [59]

Answer:

From the question we are told that

   The  population mean is  \mu  =  16

    The sample size is  n  =  30  

     The  sample mean is  \= x =  16.32

     The  population standard deviation is  \sigma  =  0.8

      The  level of significance is  \alpha  = 0.10

Step 1: State hypotheses:

The  null hypothesis is  H_o :  \mu = 16

The alternative hypothesis is  H_a :  \mu \ne  16

Step 2: State the test statistic. Since we know the population standard deviation and the sample is large our test statistics is

          t = \frac{ \= x  -\mu }{ \frac{\sigma}{ \sqrt{n} } }

=>       t = \frac{ 16.32  -16  }{ \frac{0.8 }{ \sqrt{30} } }

=>       t =2.191

Generally the degree of freedom is mathematically represented as

          df  =  n - 1

=>      df  =  30  - 1

=>      df  =29

Step 3: State the critical region(s):

From the student t-distribution table the critical value corresponding to  \alpha  = 0.10  is

         t = 1.311

Generally the critical regions is mathematically represented as  

         - 1.311 < T <  1.311

Step 4: Conduct the experiment/study:

Generally the from the value obtained we see that the t value is outside the critical region so  the decision is [Reject the null hypothesis ]  

Step 5: Reach conclusions and state in English:

  There is  sufficient evidence to show that the filling weight has to be adjusted

Step 6: Calculate the p-value associated with this test. How does this the p-value support your conclusions in Step 5?

From the student t-distribution table the probability value to the right corresponding to t =2.191 at  a degree of freedom of  df  =29  is

        P( t > 2.191) =  0.0183

Generally the p-value is mathematically represented as

       p-value  = 2 * P( t >  2.191 )

=>    p-value  = 2 * 0.0183

=>    p-value  =  0.0366

Generally  looking at the value obtained we see that p- value <  \alpha hence

The decision rule is

Reject the null hypothesis

Step-by-step explanation:

4 0
3 years ago
Solve for w what is w ÷ 1 = 6.6​
Aneli [31]

Answer:

Step-by-step explanation:

solution is w = 6.6

4 0
3 years ago
What is the cosine of the complimentary angle to 58°
Nana76 [90]

is it .119? I havent done cos, sin, tan in awhile

8 0
3 years ago
A rep hockey team has two goalies. Goalie A has a save percentage of 0.92 and Goalie B has a save percentage of 0.89. If goalie
hjlf

Answer:

a) 0.9125

b) 0.756

c) 0.314

Step-by-step explanation:

The save percentage of Goalie A = 0.92

The save percentage of Goalie B = 0.89

The percentage of the games Goalie A plays = 75 percent

The percentage of the games Goalie B plays = 25 percent

a) The probability that a save has been made by either Goalie A or Goalie B

Therefore we get;

The probability that Goalie A made a save, P(A) = 0.92 × 0.75 = 0.69

The probability that Goalie B made a save, P(B) = 0.89 × 0.25 = 0.2225

For mutually exclusive events, we have;

P(A) or P(B)  = P(A) + P(B)

∴ P(A) or P(B) = 0.69 + 0.2225 = 0.9125

The probability that a save has been made by either Goalie A or Goalie B, P(A) or P(B) = 0.9125

b) Where there are 10,000 attempts, made evenly in all games, we have;

The number of attempts made with Goalie A playing = 10,000 × 0.75 = 7,500

The number of saves Goalie A makes = 0.92 × 7,500 = 6,900

The number of saves for 10,000 attempt = (P(A) or P(B)) × 10,000 = 0.9125 × 10,000

The number of saves for 10,000 attempt = 9,125

∴ The probability that Goalie A saves = 6,900/9,125 ≈ 0.756

c) Out of the 2,500 attempts at Goalie B, the number of goals let in is given as follows;

P(Goal B) = 2,500 - 2,500 × 0.89 = 275

The number of goals let in by Goalie A, we have;

P(Goal A) = 7,500 - 7,500 × 0.92 = 600

Therefore, the total number of goals let in = 300 + 275 = 875

The probability that the goals was let in by Goalie B = 275/875 = 11/35 ≈ 0.314.

7 0
3 years ago
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