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liq [111]
2 years ago
12

I need help fast please ​

Mathematics
1 answer:
marta [7]2 years ago
5 0

Answer:

point H is in quadrant II

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Can someone help me with math!!!​
Lady bird [3.3K]

Answer:

1.  20

2.  8

3.  6

4. 25^7

5. 9^9

6.  C

Step-by-step explanation:

7 0
3 years ago
Is it possible for two numbers to have a difference of 8, and a sum of 1?
mr_godi [17]

Answer:

It depends on the type of numbers that are acceptable. If you state your problem in words

x - y =8

x + y = 1

The answers are    x = 4.5

                           y = -3.5

8 0
3 years ago
Find the values of the variables.
8_murik_8 [283]

Answer:

x = 16  y = 22.627 (or 16\sqrt{2)

Step-by-step explanation:

cos(45) = 16/y

y = 16/cos(45)

y = 22.627

tan(45) = 16/x

x = 16/tan(45)

x = 16

*this can also be found because we know this is a right triangle with two angles of 45 degrees. This means that both sides will be equal so x will be 16 and the diagonal will be 16 root 2 (22.627)

5 0
3 years ago
If c(x)= 5/x-2 and d(x) = x + 3, what is the domain of (cd)(x)?
Nataly [62]

(fg)(x)=f(x)\cdot g(x)\\\\\text{We have}\ c(x)=\dfrac{5}{x-2}\ \text{and}\ d(x)=x+3.\ \text{Substitute}\\\\(cd)(x)=\dfrac{5}{x-2}\cdot(x+3)=\dfrac{5(x+3)}{x-2}\\\\\text{Domain:}\\\\\text{The denominator must be different than zero. Therefore}\\\\x-2\neq0\to x\neq2\\\\Answer:\ \text{Domain is the set of all real numbers except 2}\to D=\mathbb{R}-\{2\}

3 0
3 years ago
At a large Midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen
guajiro [1.7K]

Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

p_1 = \frac{20}{100} = 0.2

s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04

p2 -> 1997

10 out of 100, so:

p_2 = \frac{10}{100} = 0.1

s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03

Distribution of p1 – p2:

p = p_1 - p_2 = 0.2 - 0.1 = 0.1

s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05

Confidence interval:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.  

The lower bound of the interval is:

p - zs = 0.1 - 1.645*0.05 = 0.01775


The upper bound of the interval is:

p + zs = 0.1 + 1.645*0.05 = 0.18225


The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

6 0
3 years ago
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