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3 years ago

8

Suppose that the breaking strength of a rope (in pounds) is normally distributed, with a mean of 100 pounds and a standard devia

tion of 17. What is the probability that a certain rope will break before being subjected to 130 pounds?

1 answer:

EleoNora [17]3 years ago

8 0

Answer:

0.961

Step-by-step explanation:

To answer this, find the area under the standard normal curve to the left of 130 pounds:

Using the function normalcdf( on a TI calculator, we get:

normalcdf(-1000, 130, 100, 17) = 0.961

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I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

3 years ago

What digits makes 3,71? Divisible by 3 (connections academy unit test)

wlad13 [49]

Since we know that a number is divisible by 3 if the sum of all digits is divisible by 3.

Let us check which digits will make 371? divisible by 3.

$3+7+1+0=11$

11 is not divisible by 3.

Now let us check other digits as well.

$3+7+1+1=12$

12 is divisible by 3.

$3+7+1+4=15$

15 is also divisible by 3.

$3+7+1+7=18$

18 is divisible by 3 as well.

Therefore, 1, 4 and 7 in tenth place will make our number divisible by 3 and our numbers will be 3711, 3714 and 3717.

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4 years ago

.helppppppppppp......

Lelu [443]

Step-by-step explanation:

hope you get ur answer

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3 years ago

Read 2 more answers

Also on this one. I need to try and get a good grade on math cuz I have a C on math >3

aksik [14]

165,827 and 165,229

the hundreds spot in the first number is 8 while in the second number the hundreds spot is only two

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4 years ago

Free_Kalibri [48]

Step-by-step explanation:

This is right m does equal 13. D first you multiply everything in the para thesis and so on

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3 years ago