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2 years ago

Need help with problem 5, Thank you! Sorry for the glare

Mathematics

1 answer:

MAXImum [283]2 years ago

6 0

Answer:

x = 57

Step-by-step explanation:

A triangles degrees add up to 180

67 + 56 = 123

180 - 123 = 57

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4. Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on t

MariettaO [177]

Answer:

In this case we use the Poisson distribution because we are talking about the occurrence of an event (number of tracks) over a specified interval (in this case an area interval).

The probability of the event occurring x times over an interval is:

P(x) = nˣ × e⁻ⁿ ÷ x!

where n is the mean.

a) P(7) = 6⁷ × e⁻⁶ ÷ 7! = 0.1376

b) P(x ≥ 3) = 1 - P(x < 3) = 1 - P(2) - P(1) - P(0)

P(2) = 6² × e⁻⁶ ÷ 2! = 0.0446

P(1) = 6¹ × e⁻⁶ ÷ 1! = 0.0149

P(0) = 6⁰ × e⁻⁶ ÷ 0! = 0.0025

P(x ≥ 3) = 0.9380

c) P(2 < x < 7) = P(3) + P(4) + P(5) + P(6) = 0.0892 + 0.1339 + 0.1606 + 0.1606 = 0.5443

d) The mean is going to be 6.

e) The standard deviation is √n = √6 = 2.4

3 0

2 years ago

For his off-road bike, Jackson measures that he needs a brake cable 24 1/8 in. long. The parts catalog shows cables in decimal m

statuscvo [17]

.125 * 8 = 1

A is the answer.

8 0

2 years ago

All rectangles are _____. parallelograms rhombuses squares quadrilaterals

lara31 [8.8K]

Quadrilateral and parallelogram

Squares are rectangles but rectangles are not squares b/c (not all sides are same length)

8 0

3 years ago

A gumball has a diameter that is 66 mm. The diameter of the gumball's spherical hollow core is 58 mm. What is the volume of the

grigory [225]

Answer:

Volume of gumball without including its hollow core is 48347.6 cubic mm.

Step-by-step explanation:

Given:

Diameter of Gumball = 66 mm

Since radius is half of diameter.

Radius of gumball = $\frac{diameter}{2}=\frac{66}{2} =33 \ mm$

Now We will first find the Volume of Gumball.

To find the Volume of Gumball we will use volume of sphere which is given as;

Volume of Sphere = $\frac{4}{3}\pi r^3$

Now Volume of Gumball = $\frac{4}{3}\times3.14 \times (33)^3 = 150456.24 \ mm^3$

Also Given

Diameter of gumball's spherical hollow core = 58 mm

Since radius is half of diameter.

Radius of gumball's spherical hollow core = $\frac{diameter}{2}=\frac{58}{2} =29 \ mm$

Now We will find the Volume of gumball's spherical hollow core.

Volume of Sphere = $\frac{4}{3}\pi r^3$

So Volume of gumball's spherical hollow core = $\frac{4}{3}\times3.14 \times (29)^3 = 102108.61 \ mm^3$

Now We need to find volume of the gumball without including its hollow core.

So, To find volume of the gumball without including its hollow core we would Subtract Volume of gumball spherical hollow core from Volume of Gumball.

volume of the gumball without including its hollow core = Volume of Gumball - Volume of gumball's spherical hollow core = $150456.24\ mm^3 - 102108.61\ mm^3 = 48347.63\ mm^3$

Rounding to nearest tenth we get;

volume of the gumball without including its hollow core = $48347.6\ mm^3$

Hence Volume of gumball without including its hollow core is 48347.6 cubic mm.

8 0

3 years ago

What is a real life example of trigonometric ratios

Brut [27]

Answer:

Trigonometry can be used to measure the height of a building or mountains

Step-by-step explanation:

if you know the distance from where you observe the building and the angle of elevation you can easily find the height of the building. Similarly, if you have the value of one side and the angle of depression from the top of the building you can find and another side in the triangle, all you need to know is one side and angle of the triangle.

3 0

2 years ago