52=4z HELP PLS ILL MARK BRAINLIEST

Mathematics

1 answer:

nikdorinn [45]3 years ago

5 0

Answer:

Z- 13

Step-by-step explanation:

Divide 52 by 4

$52 \div 4 =13$

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The Atlanta Braves baseball team has a mean batting average of 250 with a standard deviation of 20. Assume the batting averages

nignag [31]

Answer:

68 %

Step-by-step explanation:

Since we have our mean x = 250 and standard deviation σ = 20, we need to find how many standard deviations away the values 230 and 270 are.

Note x - σ = 250 - 20 = 230 and x + σ = 250 + 20 = 270

The values are one standard deviation away.

So, the values between 230 and 270 lie in the range x - σ to x + σ.

Since the batting averages are approximately normally distributed and for a normal distribution, 68 % of the values lie in the range x - σ to x + σ.

So, 68 % of Braves players fall between 230 and 270.

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3 years ago

Classify ABC by its angles when mA = 32°, mB = 95°, and mC = 53°.

egoroff_w [7]

Obtuse because it only takes one obtuse angle to make an obtuse triangle

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3 years ago

Read 2 more answers

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$