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Olegator [25]
3 years ago
12

Help me plzzzzzzzzzzzzzzzz

Mathematics
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

1. 25

2. 9

Step-by-step explanation:

If I'm correct, to complete the square you need to use (\frac{b}{2} )^{2}

So, for #1, b is 10. \frac{10}{2} = 5. 5^{2} = 25.

For #2, b is -6. \frac{-6}{2} = -3. -3^{2} = 9.

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Can some please answer this
Sonbull [250]

9514 1404 393

Answer:

  B  (2 violet)

Step-by-step explanation:

Let the colors be represented by their first letter. Then the balances give rise to two equation:

  o +y +b +v = r +g +b

  o +y +v = g +v

Subtracting the second equation from the first gives ...

  (o +y +b +v) -(o +y +v) = (r +g +b) -(g +v)

  b = r +b -v

Adding v-b to both sides, we have ...

  b +(v -b) = r +b -v +(v -b)

  v = r

Multiplying this result by 2, we can match the final balance:

  2r = 2v . . . . . . . choice B

7 0
3 years ago
Read 2 more answers
What is the number in scientific notation? ​0.0765
devlian [24]

Answer:

7.65 * 10^{-2} is scientific notation of 0.0765

Step-by-step explanation:

For scientific notation , follow the following notation:

a * 10^{b} -------->(equation 1)

Let,  a^{'} =  0.0765

In standard form

 a =  7.65

As decimal point move 2 places from left to right, therefore

b = -2

Put the value of 'a' and 'b' in (equation 1)

7.65 * 10^{-2} is scientific notation of 0.0765

4 0
4 years ago
The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0
valina [46]

Answer:

a) P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)

And using the complement rule, a calculator, excel or the normal standard table we have that:

P(Z>0.707)=1-P(Z

b) z=-0.674

And if we solve for a we got

a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

c) P( \bar X >1.305) = 0.05  

We can use the z score formula:

P( \bar X >1.305) = 1-P(\bar X

Then we have this:

P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

Let X the random variable that represent the amount of warpage of a population and we know

Where \mu=1.3 and \sigma=0.1

Since the sample size is large enough we can use the central limit theorem andwe know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

We can find the probability required like this:

P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)

And using the complement rule, a calculator, excel or the normal standard table we have that:

P(Z>0.707)=1-P(Z

Part b

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.75   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for a we got

a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

Part c

For this case we want this condition:

P( \bar X >1.305) = 0.05  

We can use the z score formula:

P( \bar X >1.305) = 1-P(\bar X

Then we have this:

P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076

6 0
3 years ago
What is 46,491 rounded to the nearest ten thousand
postnew [5]
50,000 is the nearest ten thousand.
8 0
3 years ago
Read 2 more answers
How do u find rate of change?
Irina18 [472]
1/3 is the rate of change. the rate of change is the slope of an equation
4 0
3 years ago
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