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3 years ago

Y=X+ ??? plz answerrr

Mathematics

1 answer:

Paul [167]3 years ago

4 0

Answer:

y=3 over -6x + 5

Step-by-step explanation:

put the 3 and -6 in a fraction

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Which of the following ratios is equivalent to 2:3 A) 1/2 B) 4/6 C) 12/13 D)20/25

kicyunya [14]

Answer:

B:4/6

Step-by-step explanation:

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3 years ago

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The surface area of this cube is 96 square meters. what is the value of y?

ELEN [110]

The formula for finding the surface area of a cube is A=6y^2 where y is the side length. we know a=96. so we just isolate y.
96/6= 6y^2 / 6
16=y^2
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4=y

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4 years ago

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The Magazine Mass Marketing Company has received 10 entries in its latest sweepstakes. They know that the probability of receivi

Serjik [45]

Answer:

P(X>4)= 0.624

Step-by-step explanation:

Given that

n = 10

p= 0.5 ,q= 1 - p = 0.5

Two fifth of 10 = 2/5 x 10 =4

It means that we have to find probability P(X>4).

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

We know that

$P(X=x)=(_{x}^{n})\ p^xq^{n-x}$

$P(X=0)=(_{0}^{10})\ 0.5^0\ 0.5^{10}=0.0009$

$P(X=1)=(_{1}^{10})\ 0.5^1\ 0.5^{9}=0.0097$

$P(X=2)=(_{2}^{10})\ 0.5^2\ 0.5^{8}=0.043$

$P(X=3)=(_{3}^{10})\ 0.5^3\ 0.5^{7}=0.117$

$P(X=4)=(_{4}^{10})\ 0.5^3\ 0.5^{7}=0.205$

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

P(X>4)= 1 -0.0009 - 0.0097 - 0.043 - 0.117-0.205

P(X>4)= 0.624

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3 years ago

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Equation by elimination -4x-9y=20 -2x-9y=-24

storchak [24]

I'm not 100% sure this is right but I think this is how you do it

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3 years ago

A contractor is required by a county planning department to submit one, two, three, four, five, or six forms (depending on the n

Ratling [72]

Answer:

$k = \dfrac{1}{21}$

b) 0.476

c) 0.667

Step-by-step explanation:

We are given the following in the question:

Y = the number of forms required of the next applicant.

Y: 1, 2, 3, 4, 5, 6

The probability is given by:

$P(y) = ky$

a) Property of discrete probability distribution:

$\displaystyle\sum P(y_i) = 1\\\\\Rightarrow k(1+2+3+4+5+6) = 1\\\\\Rightarrow k(21) = 1\\\\\Rightarrow k = \dfrac{1}{21}$

b) at most four forms are required

$P(y \leq 4) = \displaystyle\sum^{y=4}_{y=1}P(y_i)\\\\P(y \leq 4) = \dfrac{1}{21}(1+2+3+4) = \dfrac{10}{21} = 0.476$

c) probability that between two and five forms (inclusive) are required

$P(2\leq y \leq 5) = \displaystyle\sum^{y=5}_{y=2}P(y_i)\\\\P(2\leq y \leq 5) = \dfrac{1}{21}(2+3+4+5) = \dfrac{14}{21} = 0.667$

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3 years ago

Y=__X+__ ??? plz answerrr

Y=X+ ??? plz answerrr