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azamat
3 years ago
11

Which fraction is not equivalent to 7 /30

Mathematics
2 answers:
dybincka [34]3 years ago
8 0
These are the ones that are not equivalent to eachother. 1. 18/30
2. 8/10
3. 4/9
4. 14/18
5. 10/20
6. 20/32
7. 18/30

RideAnS [48]3 years ago
8 0

Answer:

6

Step-by-step explanation:

its is not step by step if u move the decimal to the right of the book

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Suppose x has a distribution with μ = 40 and σ = 12. (a) If random samples of size n = 16 are selected, can we say anything abou
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Answer:

a: no the sample size is too small

b:  Yes, the distribution is normal with a mean of 40 and standard deviation of 12

Step-by-step explanation:

a:  If n < 30, we need to know that the sample is normally distributed or else we can't determine anything.  When sample sized get very large, they usually resemble normally distributed data sets so we can still make conjectures even if the data isn't officially normally distributed

b: The question tells us that the sample is normally distributed, so even though n < 30, we can still make conjectures about the population

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3 years ago
An isosceles triangle has base angles that each measure 42 degrees.Which equation can be used to find z, the measure of the thir
gavmur [86]

Answer:

z+42+42=180

Step-by-step explanation:

Since it is an isosceles triangle, the two base angles are the same and remember all the angles add to  180. So use this equation:

z+42+42=180

If you need to find z, it is 96.

Check: 96+42+42=180

3 0
3 years ago
A function f(x) is graphed. What is the slope of the function? m= What is the y-intercept of the function? b= Which equation rep
timofeeve [1]

Answer:

slope is m = 2

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
When finding the 9th term in a geometric sequence with a common ratio of 2 and a first
Zarrin [17]

Answer:

8

Step-by-step explanation:

Nth term of a geometric sequence is given as:

t_n = a r^{(n-1)} ...(1)

Plugging n = 9, a = 3 and r = 2 in the above equation, we find:

t_9 = (3) (2)^{9-1}

t_9 = (3) (2)^{8}... (2)

Comparing equations (1) & (2), we obtan:

(n - 1) = 8

5 0
3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
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