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tekilochka [14]
3 years ago
14

Angelo's Pizza is having a ticket raffle to raise money for new soccer uniforms. A $1 ticket gives an even chance to win $25 gif

t certificate, one $15 gift certificate, or three $5 gift certificates. What is the expected profit or loss for purchasing 1 ticket if 100 total tickets are sold?
Mathematics
1 answer:
Bingel [31]3 years ago
7 0

Answer:

1-25-15-5=44

Step-by-step explanation:

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Suppose Tommy walks from his home at (0, 0) to the mall at (0, 8), and then walks to a movie theater at (6, 8). After leaving th
stiv31 [10]

Answer:

He walks 28 blocks.

Step-by-step explanation:

0,0 to 0,8 is 8 blocks

0,8 to 6,8 is 6 blocks

6,8 to 6,0 is 8 blocks

returning home (6,0 to 0,0) is 6 blocks

8+6+8+6=28 blocks

6 0
4 years ago
brigham bought a soda for $1.25 nancy bought a soda for $1.99 how much more did nancy pay for her soda
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Nancy's soda was 74 cents ($0.74) mo0re than Brigham's soda.

8 0
4 years ago
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What 2 numbers make 30.6
soldi70 [24.7K]

i could give you many

30 + 0.6

15.3 + 15.3

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3 0
3 years ago
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Someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi. We think the
denpristay [2]

Answer:

The sample size must be greater than 37 if we want to reject the null hypothesis.

Step-by-step explanation:

We are given that someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi.

Also, we are given a level of significance of 5%.

Let \mu = <u><em>mean breaking strength of their climbing rope</em></u>

SO, Null Hypothesis, H_0 : \mu = 2,000 psi       {means that the mean breaking strength of their climbing rope is 2,000 psi}

Alternate Hypothesis, H_A : \mu < 2,000 psi      {means that the mean breaking strength of their climbing rope is lower than 2,000 psi}

Now, the test statistics that we will use here is One-sample z-test statistics as we know about population standard deviation;

                         T.S.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = ample mean strength = 1,997.2956 psi

            \sigma = population standard devaition = 10 psi

            n = sample size

Now, at the 5% level of significance, the z table gives a critical value of -1.645 for the left-tailed test.

So, to reject our null hypothesis our test statistics must be less than -1.645 as only then we have sufficient evidence to reject our null hypothesis.

SO,  T.S. < -1.645   {then reject null hypothesis}

         \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < -1.645

         \frac{1,997.2956-2,000}{\frac{10}{\sqrt{n} } } < -1.645

         (\frac{1,997.2956-2,000}{10}) \times {\sqrt{n} } } < -1.645

          -0.27044 \times \sqrt{n}< -1.645

               \sqrt{n}> \frac{-1.645}{-0.27044}

                 \sqrt{n}>6.083

                  n > 36.99 ≈ 37.

SO, the sample size must be greater than 37 if we want to reject the null hypothesis.

7 0
4 years ago
Active
Naily [24]

Answer:ep 1: Put the numbers in order. ...

Step 2: Find the median. ...

Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. ...

Step 4: Find Q1 and Q3. ...

Step 5: Subtract Q1 from Q3 to find the interquartile range

Step-by-step explanation:

8 0
3 years ago
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