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ziro4ka [17]
3 years ago
10

A company introduced a much smaller variant of its tablet, known as the tablet junior. Weighing less than 11 ounces, it was abou

t 50% lighter than the standard tablet. Battery tests for the tablet junior showed a mean life of 10.25 hours. Assume that battery life of the tablet junior is uniformly distributed between 8.5 and 12 hours. What is the probability that the battery life for a tablet junior will be 9.5 hours or less?
Mathematics
1 answer:
Inessa [10]3 years ago
7 0

Answer:

0.2857

Step-by-step explanation:

Given that a company introduced a much smaller variant of its tablet, known as the tablet junior.

The weight was less than 11 oz. i.e. 50% lighter than std.

Let X be the battery life of the tablet junior

Then X is U(8.5,12)

Mean of X =10.25 hrs

the probability that the battery life for a tablet junior will be 9.5 hours or less

=P(X\leq 9.5)\\=F(9.5)\\=\frac{9.5-8.5}{12-8.5} \\=0.2857

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Answer:

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Step-by-step explanation:

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3 years ago
Approximately how long is a leg of an isosceles right triangle that has a hypotenuse of length 11.31
Lapatulllka [165]

Answer:

Leg of an isosceles right triangle is 7.99 long.

Step-by-step explanation:

Given:

Length of the hypotenuse =11.31

To find:

Length of the leg of an isosceles right triangle =?

Solution:

According to Pythagorean's Theorem, we have

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So equation 1 can be rewritten as

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2a^2 = c^2

Substituting the value of hypotenuse

2a^2 = (11.31)^2

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a = \sqrt{63.955}

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3 years ago
How do you solve this triangle by using the law of sine where
mars1129 [50]
Actually, when you know 2 sides and an included angle, you use the Law of Cosines. (and we don't know if theta is an included angle).
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c^2 = 36 + 16 - 2*6*4 * cos(60)
c^2 = 52 -48*.5
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Using the Law of Sines
side c / sin(C) = side b / sin (B)
5.2915 / sin(60) = 4 / sin (B)
sin(B) = sin(60) * 4 / 5.2915
sin(B) = 0.86603 * 4 / 5.2915
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Angle B = 40.894 Degrees

sin (A) / side a = sin (B) / side b
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sin (A) = 6 * 0.65466 / 4
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angle A = 79.109 Degrees

angle C = 60 Degrees


7 0
4 years ago
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