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3 years ago

Given x^2 + y^2 = 36 and xy = -10 find x + y.

1 answer:

8 0

This looks fun

ok so
we will use subsitution
xy=-10
divide both sides by x or y (I will choose y)
x=-10/y

sub -10/y fo x

(-10/y)^2+y^2=36
100/(y^2)+y^2=36
times both sides by y^2
100+y^4=36y^2
minus 36y^2 from both sides
y^4-36y^2+100=0
(y^2)^2-36(y^2)+100=0
quadratic formula
for
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
x= $\frac{-(-36)+/- \sqrt{(-36)^2-4(1)(100)} }{2(1)}$
x= $18+/- 4\sqrt{14}$

sub
y=-10/x
y= $\frac{-10}{18+/- 4\sqrt{14}}$

so x+y= $18+/- 4\sqrt{14}$ + $\frac{-10}{18+/- 4\sqrt{14}}$
multiply first number by $\frac{18+/- 4\sqrt{14}}{18+/- 4\sqrt{14}}$ and add them

x+y= $\frac{(18+/- 4\sqrt{14})(18+/- 4\sqrt{14})-10}{18+/- 4\sqrt{14}}$
or
x+y= $\frac{81+22 \sqrt{14} }{5}$ or $\frac{81-22 \sqrt{14} }{5}$

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snow_lady [41]

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3 years ago

IQ scores based on the Stanford-Binet IQ test are normally distributed with a mean of 100 and standard deviation 15. If you wer

lianna [129]

Answer:

95 intervals

Step-by-step explanation:

Given that

population mean = 100

standard deviation = 15

number of interval that involve the mean for 95% confidence interval is calculated as

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similarly

Assuming we measure a confidence interval of 95 percent, then we should expect 95 out of 100 confidence interval to provide the mean population therefore, answer is 95

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3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago