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Morgarella [4.7K]
3 years ago
15

Please help I will give Brantley if right Question 1 (5 points)

Mathematics
1 answer:
Vlad1618 [11]3 years ago
6 0
The answer is 4) 60miles per 4 gallons
Explanation: 60:4=15 so it will be 15 miles per 1
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How many times can 80 go into 103
Black_prince [1.1K]
1 with a remainder of 23
7 0
4 years ago
Read 2 more answers
F(x) = -4x + 5 and g(x) = x2 + 2<br> Find (f + g)(x)
xxTIMURxx [149]

Step-by-step explanation:

please mark me as brainlest

7 0
2 years ago
7 measured data points have a sample mean of 1403 and a standard deviation of 27. Determine the best estimate of the mean value
mina [271]

Answer:

Step-by-step explanation:

Hello!

You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.

Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):

[X[bar]±t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{n} }]

t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365

[1403±2.365*\frac{27}{\sqrt{7} }]

[1378.87;1427.13]

The margin of error is the semiamplitude of the interval and you can calculate it as:

d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.

I hope it helps!

4 0
3 years ago
Help me and I’ll give you brainliest
Delvig [45]

Answer:

49

Step-by-step explanation:

5 0
3 years ago
Evaluate the given integral by changing to polar coordinates.
Romashka-Z-Leto [24]

In polar coordinate, <em>R</em> is the set of points

{(<em>r</em>, <em>θ</em>) | 1 < <em>r</em> < 5 and 0 < <em>θ</em> < <em>π</em>/2}

So the integral is

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA = \int_0^{\frac\pi2}\int_1^5 r\sin(r^2)\,\mathrm dr\,\mathrm d\theta

=\displaystyle\frac\pi2\int_1^5 r\sin(r^2)\,\mathrm dr

=\displaystyle\frac\pi4\int_1^5 2r\sin(r^2)\,\mathrm dr

=\displaystyle\frac\pi4\int_1^{25} \sin(s)\,\mathrm ds

(where <em>s</em> = <em>r</em> ²)

=\displaystyle-\frac\pi4\cos(s)\bigg|_1^{25}= \boxed{\dfrac\pi4 (\cos(1) - \cos(25))}

3 0
3 years ago
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