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melomori [17]
3 years ago
9

How many equilateral triangles are there in a regular hexagon?A. SixB. FiveC. ThreeD. Eight

Mathematics
2 answers:
vesna_86 [32]3 years ago
7 0
There'd be 6 EQUILATERAL triangles in a REGULAR hexagon

faust18 [17]3 years ago
6 0

Answer:

It’s six

Step-by-step explanation:


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Neporo4naja [7]

It would be 80 gallons of water that the dishwasher will use to wash 10 loads of dishes. To figure this out you would need to divide 32 by 4 which would equal 8. You would do this to figure out how many gallons of water a dishwasher needs to wash 1 load of dishes. Then you would multiply 8 times 10 which would equal 80. Hope this helps you!

8 0
3 years ago
Please get me the answer
Anna71 [15]

Answer:

i belive its A $15

Step-by-step explanation:

if $3 is 20% and we are trying to find the full 100% then you miltiply 3 by 5 since 20 times 5 equals 100 and when you multiply 3 by 5 you get 15 dollars which is 100 percent of the iteam meaning the regular price.

3 0
2 years ago
Write each as a decimal.Round to the thousands place​
max2010maxim [7]

Answer:

2) 0.3

4) 0.09

6) 0.65

8) 4.45

Step-by-step explanation:

hope this helps ;)

7 0
3 years ago
Read 2 more answers
This problem uses the teengamb data set in the faraway package. Fit a model with gamble as the response and the other variables
hichkok12 [17]

Answer:

A. 95% confidence interval of gamble amount is (18.78277, 37.70227)

B. The 95% confidence interval of gamble amount is (42.23237, 100.3835)

C. 95% confidence interval of sqrt(gamble) is (3.180676, 4.918371)

D. The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

Step-by-step explanation:

to)

We will see a code with which it can be predicted that an average man with income and verbal score maintains an appropriate 95% CI.

attach (teengamb)

model = lm (bet ~ sex + status + income + verbal)

newdata = data.frame (sex = 0, state = mean (state), income = mean (income), verbal = mean (verbal))

predict (model, new data, interval = "predict")

lwr upr setting

28.24252 -18.51536 75.00039

we can deduce that an average man, with income and verbal score can play 28.24252 times

using the following formula you can obtain the confidence interval for the bet amount of 95%

predict (model, new data, range = "confidence")

lwr upr setting

28.24252 18.78277 37.70227

as a result, the confidence interval of 95% of the bet amount is (18.78277, 37.70227)

b)

Run the following command to predict a man with maximum values ​​for status, income, and verbal score.

newdata1 = data.frame (sex = 0, state = max (state), income = max (income), verbal = max (verbal))

predict (model, new data1, interval = "confidence")

lwr upr setting

71.30794 42.23237 100.3835

we can deduce that a man with the maximum state, income and verbal punctuation is going to bet 71.30794

The 95% confidence interval of the bet amount is (42.23237, 100.3835)

it is observed that the confidence interval is wider for a man in maximum state than for an average man, it is an expected data because the bet value will be higher than the person with maximum state that the average what you carried s that simultaneously The, the standard error and the width of the confidence interval is wider for maximum data values.

(C)

Run the following code for the new model and predict the answer.

model1 = lm (sqrt (bet) ~ sex + status + income + verbal)

we replace:

predict (model1, new data, range = "confidence")

lwr upr setting

4,049523 3,180676 4.918371

The predicted sqrt (bet) is 4.049523. which is equal to the bet amount is 16.39864.

The 95% confidence interval of sqrt (wager) is (3.180676, 4.918371)

(d)

We will see the code to predict women with status = 20, income = 1, verbal = 10.

newdata2 = data.frame (sex = 1, state = 20, income = 1, verbal = 10)

predict (model1, new data2, interval = "confidence")

lwr upr setting

-2.08648 -4.445937 0.272978

The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

4 0
3 years ago
The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.9 mg and standard deviation 0.1
Ad libitum [116K]

Answer:

The answer to the question is;

The probability that the resulting sample mean of nicotine content will be less than 0.89 is 0.1587 or 15.87 %.

Step-by-step explanation:

The mean of the distribution = 0.9 mg

The standard deviation of the sample = 0.1 mg

The size of the sample = 100

The mean of he sample = 0.89

The z score for sample mean is given by

Z =\frac{X-\mu}{\sigma/ \sqrt{n} } where

X = Mean of the sample

μ = Mean of the population

σ = Standard deviation of the population

Therefore Z = \frac{0.89-0.90}{0.1/\sqrt{100} } = -1

From the standard probabilities table we have the probability for  a z value of -1.0 = 0.1587

Therefore the probability that the resulting sample mean will be less than 0.89 = 0.1587 That is the probability that the mean is will be less than 0.89 is 15.87 % probability.

             

7 0
3 years ago
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