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MAVERICK [17]
3 years ago
14

Can I get help please

Mathematics
1 answer:
Arisa [49]3 years ago
6 0
Miles is going to spend fewer hours bicycling this week than he did last year.
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"a man drove 11 mi directly east from his home, made a left turn at an intersection, and then traveled 2 mi north to his place o
Charra [1.4K]
Pythagorean therom!


east and the north make a right angle

a^2 + b^2 = c^2
(11)^2 + (2)^2 = c^2
121 + 4 = c^2
125 = c^2
11.2 miles = c
8 0
3 years ago
Jenna ordered 28 shirts for her soccer team 75% of those shirts size large how many large shirts did you know order
Ymorist [56]
28 x 75% = 21
Jenna ordered 21 large shirts.
8 0
3 years ago
What is α, the probability of a Type I error if the null hypothesis, H0, is: Carmin believes that her chemistry exam will only c
baherus [9]

Answer:

The probability of falling into a type I error, when testing a hypothesis test, consists of:

Probability of rejecting the null hypothesis when, in reality, this hypothesis is true.

Probability of rejecting the null hypothesis when, in reality, this hypothesis is true, is:

Probability of Affirm that Chemistry exam will NOT cover only chapters four and five, since the Chemistry exam will cover only chapters four and five.

That is, alpha is the probability that Carmin decides to study additional chapters, unnecessarily.

Step-by-step explanation:

8 0
3 years ago
Someone please help me! Consider the following data set: 3, 4, 6, 7, 9, 9, 11
Korolek [52]
The range is 11- 3 = 8  Adding 5 would not affect the range.
4 0
3 years ago
A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of size 2​, drawn from the 11​, will
svet-max [94.6K]

The required probability is \frac{36}{55}

<u>Solution:</u>

Given, a shipment of 11 printers contains 2 that are defective.  

We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.

Now, we know that, \text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}

Probability for first draw to be non-defective =\frac{11-2}{11}=\frac{9}{11}

(total printers = 11; total defective printers = 2)

Probability for second draw to be non defective =\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}

(printers after first slot = 10; total defective printers = 2)

Then, total probability =\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}

7 0
3 years ago
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