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2 years ago

NEED ASAP PLS HELP

Mathematics

1 answer:

Nastasia [14]2 years ago

5 0

The sine ratio that was applied in step 2 is incorrect, therefore, the first error is in: B. step 2.

<h3>How to Apply the Sine Ratio?</h3>

The sine ratio that is applied when we want to solve a right triangle is, sin ∅ = opposite length / hypotenuse length.

In step 2, the sine ratio applied is wrong because:

∅ = W
Opposite length = XZ
Hypotenuse length = y

Thus, we should have:

sin(W) = XZ/y

The first error is in: B. step 2.

Learn more about the sine ratio on:

brainly.com/question/2920412

#SPJ1

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The probability that a student has a Visa card (event V) is .63. The probability that a student has a MasterCard (event M) is .1

Aleksandr-060686 [28]

Answer:

a) The probability that a student has either a Visa card or a MasterCard is 0.71.

b) V and M are not independent.

Step-by-step explanation:

Given : The probability that a student has a Visa card (event V) is 0.63. The probability that a student has a MasterCard (event M) is 0.11. The probability that a student has both cards is 0.03.

To find :

a) The probability that a student has either a Visa card or a MasterCard ?

b) In this problem, are V and M independent ?

Solution :

The probability that a student has a visa card(event V) is P(V)= 0.63

The probability that a student has a MasterCard (event M) is P(M)= 0.11

The probability that a student has both cards is $P(V \cap M)=0.03$

a) Probability that a student has either a Visa card or a Master Card is given by,

$P(V \cup M) = P(V) + P(M) - P(V\cap M)$

$P(V \cup M) = 0.63+ 0.11- 0.03$

$P(V \cup M) =0.74- 0.03$

$P(V \cup M) =0.71$

The probability that a student has either a Visa card or a MasterCard is 0.71.

b) Two events, A and B, are independent if $P(A\cap B)=P(A)P(B)$

For V and M to be independent the condition is satisfied,

$P(V\cap M)=P(V)P(M)$

Substitute the values,

$0.03=0.63\times 0.11$

$0.03\neq 0.0693$

So, V and M are not independent.

6 0

3 years ago

A projectile is fired with muzzle speed 220 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

Allushta [10]

Answer:

4968.6 m from where it was fired
221.33 m/s

Step-by-step explanation:

For the purpose of this problem, we assume ballistic motion over a stationary flat Earth under the influence of gravity, with no air resistance.

We can divide the motion into two components, one vertical and one horizontal. For muzzle speed s and launch angle θ, the horizontal speed is presumed constant at s·cos(θ). The initial vertical speed is then s·sin(θ) and the (x, y) coordinates as a function of time are ...

(x, y) = (s·cos(θ)·t, -4.9t² +s·sin(θ)·t + h₀) . . . . . where h₀ is the initial height

To find the range, we can solve the equation y=0 for t, and use this value of t to find x.

Using the quadratic formula, we find t at the time of landing to be ...

t = (-s·sin(θ) - √((s·sin(θ))²-4(-4.9)(h₀)))/(2(-4.9))

t = (s/9.8)(sin(θ) +√(sin(θ)² +19.6h₀/s²))

For s = 220, θ = 45°, and h₀ = 30, the time of flight is ...

t ≈ 31.939 seconds

Then the horizontal travel is

x = 220·cos(45°)·31.939 ≈ 4968.6 . . . . meters

As it happens, the value under the radical in the above expression for time, when multiplied by s, is the vertical speed at landing. The horizontal speed remains s·cos(θ), so the resultant speed is the Pythagorean sum of these:

landing speed = s·√(cos(θ)² +sin(θ)² +19.6h₀/s²) ≈ s√(1 +0.012149)

≈ 221.33 m/s

_____

Note that the landing speed represents the speed the projectile has as a consequence of the potential energy of its initial height being converted to kinetic energy that adds to the kinetic energy due to its initial muzzle velocity.

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3 years ago

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Triangle QRS is plotted in Quadrant I. After a 180° rotation about the origin, which quadrant would its image be? quadrant III

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We have the formula A = (h x b) / 2, where A is the area of the triangle, h is the height of the triangle, b is the base of the triangle;
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