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4 years ago

1.3(8-b)+3.7b= -5.2 solve and check answer

Mathematics

2 answers:

Paraphin [41]4 years ago

6 0

Answer:

b = -6.5

Step-by-step explanation:

I am 100% sure thts the right answer

Dafna1 [17]4 years ago

3 0

Answer:

b=-6.5 is your answer.

Step-by-step explanation:

What you have to do is get b by itself.

1.3(8-b)+3.7b=-5.2

10.4-1.3b+3.7b=-5.2

2.4b+10.4=-5.2

2.4b=-15.6

b=-6.5 is your answer.

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HELP! USE EITHER LAW OF SINE OR LAW OF COSINE! HELP PLEASE! FIND THE MISSING SIDE AND MISSING ANGLES!

Eva8 [605]

Answer:

∠Q ≈ 42.969, ∠P ≈ 52.03, QR ≈ 15.036

Step-by-step explanation:

Law of sines:

13/sin Q = 19/sin 85°

=> sin Q = 13 × sin 85°/19

=> Q = sin^-1 (13 × sin 85°/19)

=> Q ≈ 42.969

∠P=> 180 - 85 - 42.969 ≈ 52.03

QR/sin 52.03° = 19/sin 85°

=> QR = sin 52.03° × 19/sin 85°

=> QR ≈ 15.036

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3 years ago

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Two lines intersect to form the angles shown.

yawa3891 [41]

The answer i think it is ,is b

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3 years ago

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Surface-finish defects in a small electric appliance occur at random with a mean rate of 0.3 defects per unit. find the probabil

mario62 [17]

Answer: Probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

Step-by-step explanation:

Since we have given that

Mean rate defects per unit = 0.3

Since we will use "Poisson distribution":

$P(X=K)=\frac{e^{-\lambda}\lambda^k}{k!}$

But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.

$P(X\geq 2)=1+P(X=0)+P(X=1)$

So,

$P(X=0)=\frac{e^{-0.3}0.3^0}{0!}=e^{-0.3}=0.74\\\\P(X=1)=\frac{e^{-0.3}\times 0.3}{1}=0.22$

so, it becomes,

$P(X>2)=1-(0.74+0.22)=1-0.96=0.04$

Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

4 0

4 years ago

The length s(t)s(t)s, (, t, )of the side of the base of a square prism is decreasing at a rate of 777 kilometers per minute and

gavmur [86]

Answer:

$\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}$

Step-by-step explanation:

The surface area of the square prism is obtained by using the following formula:

$A_{s} (t) = 4\cdot l(t)\cdot h(t) + 2\cdot [l(t)]^{2}$

The rate of change of the surface area can be found by deriving the function with respect to time:

$\frac{dA_{s}}{dt} = 4\cdot [h(t)\cdot \frac{dl}{dt} + l(t)\cdot \frac{dh}{dt}] + 2\cdot l(t)\cdot \frac{dl}{dt}$

Known variables are summarized below:

$h(t) = 9\,km$

$l(t) = 4\,km$

$\frac{dh}{dt} = 10\,\frac{km}{min}$

$\frac{dl}{dt} = -7\,\frac{km}{min}$

The rate of change is:

$\frac{dA_{s}}{dt} = 4\cdot [(9\,km)\cdot (-7\,\frac{km}{min} )+(4\,km)\cdot (10\,\frac{km}{min} )] + 2\cdot (4\,km)\cdot (-7\,\frac{km}{min} )$

$\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}$

3 0

3 years ago

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Use the distance formula distance, to find the distance to the nearest tenth , between each pair of points A(6,2) and D(-3,-2)

marusya05 [52]

Answer:

The distance between A and D to the nearest tenth is;

$9.8\text{ units}$

Explanation:

Given the two points;

$A(6,2)\text{ and D(-3,-2)}$

Applying the distance between two points formula;

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

substituting the given coordinates we have;

$\begin{gathered} AD=\sqrt[]{(-3-6)^2+(-2-2)^2} \\ AD=\sqrt[]{(-9)^2+(-4)^2} \\ AD=\sqrt[]{81+16} \\ AD=\sqrt[]{97} \end{gathered}$

Simplifying;

$\begin{gathered} AD=9.8488578 \\ AD\approx9.8 \end{gathered}$

Therefore, the distance between A and D to the nearest tenth is;

$9.8\text{ units}$

4 0

2 years ago