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3 years ago

9

pls answer ive asked this question three times now pls answer everything there are five problems smh i neeeeeeeeeeeeeeeeeeeeeeee

eeeeeeeeed it bad T~T

1 answer:

mafiozo [28]3 years ago

6 0

Answer:

Ex1. X=-1

Ex2. 3rd choice

Step-by-step explanation:

Ex1. The only discontinuity in the domain is x=-1

Ex2. You just add values of x to those options and see which one applies

You might be interested in

Pretty pls help me with A and B with work SHOWN

-BARSIC- [3]

A. 11/30 x 50= 18/36=1/2

18/36 x 1/10= 5/100

B. 1,210 x 1.1= 1,331

I am not so sure about A, but I hope this helps a little.

7 0

4 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

Can someone help me find X of this triangle? thanks :)

tino4ka555 [31]

Since it is an equilateral triangle, all of the sides are the same length.

So, x=14 :)

Hope this helps!

7 0

3 years ago

Read 2 more answers

Follow the steps below to find the mean for the set of data. Use the work shown to help you. Data set: 6, 5, 8, 5, 9, 6, 7, 5, 1

lbvjy [14]

Answer:

7

Step-by-step explanation: 6 + 5 + 8 + 5 + 9 + 6 + 7 + 5 + 12 = 63 so now u just divide that by 9

6 0

3 years ago

Read 2 more answers

rewrite the polynomial 2y^2+ 6y^3-11-17y^4+8y^5 in the standard form also find its degree and coefficient of y^4

OleMash [197]

Answer:

The standard form is 8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11

The degree of given polynomial is '5'

the co-efficient of y⁴ is '-17'

Step-by-step explanation:

Given standard form 2 y²+ 6 y³-11-17 y⁴+8 y⁵

<em>The form ax² + b x + c is called the standard form of the quadratic expression of 'x'.This is second degree standard form of polynomial.</em>

<em>The form ax⁵ + b x⁴ + c x³ +d x² +ex +f is called the standard form of the quadratic expression of 'x'.This is fifth degree standard form of polynomial</em>

now Given polynomial is 2 y²+ 6 y³-11-17 y⁴+8 y⁵

The standard form is

8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11

<u><em>Conclusion</em></u>:-

<em>The degree of given polynomial is '5'</em>

<em>The co-efficient of y⁴ is '-17'</em>

<em> </em>

5 0

3 years ago