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2 years ago

Does anyone know the formula for the volume of a oblique hexagonal prism?

Mathematics

1 answer:

Scorpion4ik [409]2 years ago

8 0

Answer:

, volume = [(3√3)/2]a2h cubic units

Step-by-step explanation:

The formula for the volume of a hexagonal prism is, volume = [(3√3)/2]a2h cubic units where a is the base length and h is the height of the prism.

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PLEASEEEE HELPPP Angle A and angle B are two angles that are complementary. We know angle A measures (x+2)° and Angle B measures

saul85 [17]

Answer:

Step-by-step explanation:

x + 2 + 57 = 90

x + 59 = 90

x = 31°

x + 2 = 33°

3 0

3 years ago

Read 2 more answers

PLEASE SOMEONE HELP ME WITH THESE QUESTIONS I'VE BEEN STUCK ON HERE FOR SO LONG

Neporo4naja [7]

Answer:

# means number

Step-by-step explanation:

Distance = Y-axis or output

Cost= X-axis or input

you do not need to add the brackets for the formula(BEDMAS), but it makes things clearer

For A -

the Y-intercept is where the first coordinate, or the # or the X-axis, is 0 (on the graph the y-intercept 0,4)

For B -

y = mx + b

m = slope

x = input or # on the X-axis

b = y-intercept

formula = (input x increase in the slope) + output for 0 or (distance x increase in cost) + output for 0

input(x axis) output(y axis)

0 5 ← output for 0

1 9

2 13

there is a +4 increase in the slope, the increase in the slope is an output #(in this case 9) - the previous output number (in this case 5)

formula = (input x increase in the slope) + output for 0 = distance

formula in use = (3 x 4) + 5 = 17

For C:

The dependent variable is the output or the cost because the input or the distance will determine the cost travelled. Eg. if I travel 9 miles it will cost $41. If the dependent variable is the cost then the independent variable is the distance

For D: (input x increase in the slope) + output for 0

(4.2 miles x $4) + 5

16.8 + 5 = $21.8; It will cost $21.8 to travel 4.2miles

For E: (distance x increase in cost) + output for 0

(? x 4) + 5 = 14

(14 - 5) = 9

9 ÷ 4 = 2.25miles; with 14 dollars I could travel 2.25 miles

(It is a bit confusing lol, but this is the way I understand it, if you can't understand it this way then use y = mx + b)

4 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$