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2 years ago

Help&EXPLAIN ••••••••••••••••••• Help plz

Mathematics

1 answer:

Stolb23 [73]2 years ago

3 0

Answer:

diameter = 2.5 feet

Step-by-step explanation:

Circumference of a circle is given as C = πd

Circumference of the tree trunk = 7.85 ft

π = 3.14

Diameter of the tree trunk = d

Thus:

7.85 = 3.14*d

Divide both sides by 3.14

7.85/3.14 = 3.14*d/3.14

2.5 = d

d = 2.5 ft

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Which of the following helps bloggers decide how to present information

bekas [8.4K]

Since there is no options, i will list several reasons that help bloggers decide how to present information :
- Audience expectation
- Blogger's point of view and personal principle
- The actual fact that blogger has

8 0

3 years ago

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Write the equation of the line that passes through the points (2, 1) and (-8,-8).

nordsb [41]

Answer:

y = 0.9x - 0.8

Step-by-step explanation:

First you use the formula m = (y2-y1) / (x2-x1)

Sub in the 2 points you are given and you get m = (-8-1) / (-8-2)

m = -9/-10 = 9/10

Now u use the point slope formula, (y - y1) = m(x - x1)

Sub in the points and you get (y - 1) = 9/10 (x-2)

Expand it out to get y - 1 = 9/10x - 18/10

multiple everything by 10 = 10y - 10 = 9x -18

make y the subject, 10y = 9x - 8

y = 0.9x - 0.8

8 0

2 years ago

The derivative of y with respect to x, when y = 3e^5x - 2 is:

Ede4ka [16]

Answer: 15e^5x

Step - by - step

y=3e^5x - 2

By the sum rule, the derivative of 3e^5x - 2 with respect to x is d/dx [ 3e^5x ] + d/dx [-2].

d/dx [ 3e^5x ] + d/dx [ -2 ]

Evalute d/dx [ 3e^5x ]

Since 3 is constant with respect to x , the derivative of 3e^5x with respect to x is
3 d/dx [ e^5x ].

3 d/dx [ e^5x ] + d/dx [ -2 ]

Differentiate using the chain rule, which states that d/dx [ f(g(x))] is f' (g(x)) g' (x) where f(x) = e^x and g(x) = 5x.

To apply the Chain Rule, set u as 5x.
3 ( d/du [ e^u] d/dx [5x] ) + d/dx [ -2]

Differentiate using the Exponential rule which states that d/du [ a^u ] is a^u ln(a) where a=e.

3( e^u d/dx[5x] ) + d/dx [ -2 ]
Replace
3(e^5x d/dx [5x] ) + d/dx [ -2 ]
3(e^5x( 5 d/dx [x] )) + d/dx [ -2 ]

Diffentiate using the Power Rule which states that d/dx [x^n] is nx^n-1 where n=1.
3(e^5x(5*1)) + d/dx [-2]
3 ( e^5x * 5 ) + d/dx [-2]

Multiply 5 by 3
15e^5x + d/dx [-2]

Since -2 is constant with respect to x, the derivative of -2 with respect to x is 0.
15e^5x + 0
15e^5x

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7 months ago

The population of turtles is doubling every 3 months. If there are 400 turtles now, how many will there

QveST [7]

First we need to know how many months is 2 years. To answer that we multiply the number of months in a year by 2 and get 48 months.

Every 3 months the population of turtles doubles, but to get how many times the population doubles, you need to divide 48 months by 3 to get 16.

So population of 400 will get doubled 16 times in the following two years.

We construct the following equation where $p$ is the final population,

$p=400\cdot2\cdot16=\boxed{12800}$

Hope this helps. :)

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3 years ago

Read 2 more answers

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

2 years ago