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Illusion [34]
3 years ago
14

What’s the answer to this

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
7 0
2.4/0.3=8
0.3x8=2.4
The scale factor is 8.
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How to write 16,208.59 in word form
IrinaK [193]

Answer:

Sixteen thousand, two hundred eight and fifty-nine hundredths.

I hope this helps! :)

8 0
3 years ago
1.27 as a simplest form
jasenka [17]

Answer:

its already simplest form, if you mean fractions, then the answer is 1 27/100

Step-by-step explanation:

6 0
3 years ago
Find a unit vector in the direction of the vector left bracket Start 3 By 1 Matrix 1st Row 1st Column negative 8 2nd Row 1st Col
MrRissso [65]

Answer:

\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]

Step-by-step explanation:

We are required to find a unit vector in the direction of:

\left[\begin{array}{c}-8\\7\\2\end{array}\right]

Unit Vector, \hat{a}=\dfrac{\overrightarrow{a}}{|\overrightarrow{a}|}

The Modulus of \overrightarrow{a} =\sqrt{(-8)^2+7^2+(-2)^2}=\sqrt{117}

Therefore, the unit vector of the matrix is given as:

\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]

5 0
3 years ago
A cannon ball is launched at 6 feet per second from the top of a platform 12 feet above the ground.
AleksAgata [21]
A.) y = ut - 1/2gt^2 + 12
y = 6t - 1/2(9.8)t^2 + 12
y = 6t - 4.9t^2 + 12

b.) The y-intercept is 12

c.) The y-intercept represent the height of the ball at time t = 0

d.) At the time the ball reaches the ground, y = 0
-4.9t^2 + 6t + 12 = 0
t = 2.29 seconds
Therefore, it took the ball 2.29 seconds to reach the ground.
7 0
3 years ago
From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x ≥ 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
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