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Evgen [1.6K]
3 years ago
15

Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace

length is more than 4.03 inches
Mathematics
1 answer:
joja [24]3 years ago
4 0

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

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Please answer this for me pleaseeeee
kobusy [5.1K]

Answer:

\Huge\boxed{x=71.9}

Step-by-step explanation:

Hello There!

Once again we are going to use trigonometry to solve for x

Here are the <u>Trigonometric Ratios</u>

sin = opposite divided by hypotenuse

cos = adjacent divided by hypotenuse

tan = opposite divided by adjacent

we need to find x and we are given its opposite side length (58) and the adjacent side length (19)

this corresponds with tangent so once again we will be using tangent to solve for x

Looking at tangent we see that its equal to opposite divided by adjacent so we create an equation

tan(x)=\frac{58}{19}\\\frac{58}{19} =3.052631579

now we have

tan(x)=3.052631579

Once again we need to get rid of the tan

to do so we take the inverse of tan ( tan^-1) and apply it to each side

arcsin^-^1(tan(x))=x\\arcsin^-^1(3.052631579)=71.86191784

finally we round to the nearest tenth

we're left with x = 71.9

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Write the equation of the graphed circle.
Sveta_85 [38]
Do NOT search the link they WILL hack you
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during a softball game kay hit a fly ball the function f(x) = -16t^2 + 64t + 4 describes the height of the softball in feet. mak
STatiana [176]
<h2>Hello!</h2>

The graph is attached.

To graph a parabola we need to know the following:

- If the parabola is open upwards or downwards

- They axis intercepts (if they exist)

- The vertex position (point)

We are given the function:

f(t)=-16t^{2}+64t+4

Where,

a=-16\\b=64\\c=4

For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.

Finding the axis interception points:

Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:

f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0

Using the quadratic equation:

\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615

So, at t=4.0615 the height of the softball will be 0.

Since we will work only with positive values of "x", since we are working with a function of time:

Let's start from "t" equals to 0 to "t" equals to 4.0615.

So, evaluating we have:

f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0

Finally, we can conclude that:

- The softball reach its maximum height at t equals to 2. (68 feet)

- The softball hits the ground at t equals to 4.0615 (0 feet)

- At t equals to 0, the height of the softball is equal to 4 feet.

See the attached image for the graphic.

Have a nice day!

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Answer:

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Step-by-step explanation:

the volume is 8Mm-18M

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