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Liono4ka [1.6K]
3 years ago
13

PLZ HELP!!!!!!!!!!!!!!!

Mathematics
1 answer:
emmainna [20.7K]3 years ago
8 0

Answer:

589htuvuvfd

Step-by-step explanation:

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The weights of 1,000 men in a certain town follow a normal distribution with a mean of 150 pounds and a standard deviation of 15
galina1969 [7]
Im not really sure but 45 sir :)
7 0
4 years ago
DeShawn left the hospital and drove toward the ferry office at an average speed of 30 mph. Shaun left sometime later driving in
kvv77 [185]

Answer:

The number of hours DeShawn drove before Shaun caught up = 5 hours

Step-by-step explanation:

Average speed of DeShawn = 30 mph

Average speed of Shawn = 50 mph

After driving for three hours Shaun caught up with DeShawn.

Distance traveled by Shaun in 3 hours,

               Distance = Time x Average speed = 50 x 3 = 150 miles.

So DeShawn also travels 150 miles, we need to find time DeShawn taken to travel this 150 miles,

                Distance = Time x Average speed

                  150 = Time x 30

                   Time = 5 hours.

The number of hours DeShawn drove before Shaun caught up = 5 hours            

7 0
3 years ago
Can you help me please
Phantasy [73]
There are 8 different ways
3 0
4 years ago
Factor the linear expression 24y - 40 using the greatest common factor of the terms
ANEK [815]
Factor: 24y-40
Answer: 8(3y-5)

Simplifying: 24y-40
Answer: There are no like terms.
7 0
3 years ago
In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo
larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine z_{\frac{\alpha}{2}}=z_{0.005}

Now by using z score table we find that  z_{\frac{\alpha}{2}}=2.58

The boundaries of the confidence interval are:

\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

3 0
4 years ago
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