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3 years ago

Help me please. I don't understand this.

Mathematics

1 answer:

navik [9.2K]3 years ago

6 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

Apparently, you're being asked to identify the sequence of steps you would use to compute the volume of the pyramid.

It seems to be a good idea to start with the formula for the volume.

Then, recognize that you need to compute B, so make that computation. The area of the base (B) is the product of the base dimensions (14)(12).

Once you have the value of B, then you can put that, along with the value of h, into the original volume formula.

Evaluating it gives the volume in cubic units.

_____

Additional comment

If you're familiar with the pyramid volume formula, you know that you must compute B before you can make use of the formula. That makes the sequence be B=14(12); B=168; V=1/3Bh; V=1/3(168)(7).

However, if you're starting from scratch, it is probably good to begin with the volume formula. That is what tells you that you need to find B in the first place. This is the sequence we show below.

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At noon, ship A is 50 km west of ship B. Ship A is sailing south at 10 km/h and ship B is sailing north at 20 km/h. How fast is

Lena [83]

So hmmm check the picture below

so, we're looking for dr/dt then at 4:00pm or 4 hours later

now, keep in mind that, the distance "x", is not changing, is constant whilst "y" and "r" are moving, that simply means when taking the derivative, that goes to 0

$\bf r^2=x^2+y^2\implies 2r\cfrac{dr}{dt}=0+2y\cfrac{dy}{dt}\implies \cfrac{dr}{dt}=\cfrac{y\frac{dy}{dt}}{r}\quad 
\begin{cases}
\cfrac{dy}{dt}=30\\
r=130\\
y=120
\end{cases}
\\\\\\
\cfrac{dr}{dt}=\cfrac{120\cdot 30}{130}$

6 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

Scott wants to cover his patio with concrete pavers. If the patio is shaped like a trapezoid whose bases are 11 feet and 13 feet

GenaCL600 [577]

Answer:

Scott will need 168 ft² of pavers to cover his patio

Explanation:

Scott wants to cover a trapezoid-shaped patio

This means that:

To get the number of square feet of pavers he'll need, we need to get the area of his patio

Area of trapezium id calculated as follows:

$Area = \frac{base_{1}+base_{2}}{2} * height$

We are given that:

base₁ = 11 ft

base₂ = 13 ft

height = 14 ft

We now substitute with the givens to get the area as follows:

$Area = \frac{11+13}{2} * 14 = 168ft^2$

This means that:

Scott will need 168 ft² of pavers to cover his patio

Hope this helps :)

5 0

3 years ago

758x93= What is the answer and how to solve it

Ierofanga [76]

The answer is 70494.

5 0

3 years ago

convert 13542 base six to base ten. I'm in math20 at Folsom Lake College and i'm trying to figure out this question. I had most

VikaD [51]

Converting to base 10 can be done easily by writing a sort of digit expansion. In base 6, this number literally means

$13542_6=1\times6^4+3\times6^3+5\times6^2+4\times6^1+2\times6^0$

Simplifying this gives the value in base 10.

$13542_6=1296+3\times216+5\times36+4\times6+2$
$13542_6=2150_{10}$

6 0

3 years ago