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aleksley [76]
3 years ago
14

Slove the equation y+3=-y+9 on edge A)y=1 B)y=3 C)y=6 D)y=9

Mathematics
2 answers:
Romashka [77]3 years ago
5 0

Answer:

y + 3 = -y + 9

-y         -y

3 = -2y + 9

+2y  +2y

2y + 3 = 9

     -3     -3

2y = 6

y = 3

FinnZ [79.3K]3 years ago
5 0
The right answer is y=3
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Subtract 1 from +1 and 0 and then divide the negative from x and -1 which gives u with x=1
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3 years ago
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One-third of 2c subtracted frome 5 times nine
lilavasa [31]
-44.8 is the answer.
4 0
3 years ago
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Julius is buying beverages for brunch. He needs to buy a total of 5
xeze [42]
Well, he only needs 5 gallons of beverages. Because he buys two containers of each, however, that means he bought 320 gallons total (I would say). How I got my answer:

There are 8 pints in a gallon (8 x 2)
There are 4 quarts in a gallon (4 x 2)
There are 16 cups in a gallon (16 x 2)
There are 128 ounces in a gallon (128 x 2)

With that being said, we can add these to find the total amount of gallons of beverages Julius bought. 16 + 8 + 8 + 32 + 256 = 320.

Hopefully this helps!
7 0
2 years ago
The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

5 0
3 years ago
Answer....... (-1)^3×(-1)^2
8090 [49]
Hello there!

(-1)^3 * (-1)^2

Use the distributive property

= (-1)(-1)(-1)* (-1)(-1)

= (-1)^5

= -1


I hope this helps!
3 0
3 years ago
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