Write equations of parallel & perpendicu

Which numer is rational and a real number, but not an integer

Amiraneli [1.4K]

Answer:

d

Step-by-step explanation:

sory if wrong

have a great day

8 0

3 years ago

Expand each expression A ( x+1)^2 B(x-3)^2. C ( x+4)^2 D (x-1/2)^2

Elis [28]

The <em>expanded</em> expressions are shown below:

(x + 1)² = x² + 2 · x + 1
(x - 3)² = x² - 6 · x + 9
(x + 4)² = x² + 8 · x + 16
(x - 1/2)² = x² - x + 1/4
3 · (x - 5)² = 3 · (x² - 10 · x + 25) = 3 · x² - 30 · x + 75
(1/2) · (x - 2)² = (1/2) · (x² - 4 · x + 4) = (1/2) · x² - 2 · x + 2

<h3>How to expand perfect square trinomials</h3>

<em>Perfect square</em> trinomials are polynomials of grade 2 with the form (a + b)² = a² + 2 · a · b + b². In this case, we have to expand six perfect square trinomials:

a) (x + 1)² = x² + 2 · x + 1

b) (x - 3)² = x² - 6 · x + 9

c) (x + 4)² = x² + 8 · x + 16

d) (x - 1/2)² = x² - x + 1/4

e) 3 · (x - 5)² = 3 · (x² - 10 · x + 25) = 3 · x² - 30 · x + 75

f) (1/2) · (x - 2)² = (1/2) · (x² - 4 · x + 4) = (1/2) · x² - 2 · x + 2

To learn more on perfect square trinomials: brainly.com/question/385286

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3 0

3 years ago

What is the answer for<br>2c/5<8

rosijanka [135]

Hello guys I am new here and I don't know how to answer your question

6 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%