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3 years ago

Someone who can do this help me please.

Mathematics

1 answer:

lianna [129]3 years ago

5 0

Answer:

x= 2

Step-by-step explanation:

the shapes are 6 in apart

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A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs

Hitman42 [59]

Answer:

Step-by-step explanation:

Let X be the solvent present

Then given that the manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent.

X is N(101.3, 3.68)

a) $P(X>100) = P(Z>\frac{100-101.3}{3.68} =P(Z>-0.35)\\=0.6368$

Probability = 0.6368

b) $P(X>100)= 0.97$ means Z score = 1.88

X=100+1.88(s) =101.3

s=1.3/1.88=0.6914

4 0

3 years ago

ABCD is a quadrilateral such that the measures of three of its interior angles are given as B=122 degrees, C= 75 degrees, D= 51

posledela

Answer:

angle A is 112 degree.

Step-by-step explanation:

quadilateral have 4 angles .

sum of the interior angles of a quadilateral is 360 degree

Here,

angle A = be x

angle B = 122

angle C = 75

angle D = 51

Now,

angle A + angle B + angle C + angle D =360 degree

x + 122 + 75 + 51 = 360

x + 248 = 360

x = 360 - 248

x = 112

4 0

2 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$