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3 years ago

14

Please help me !!!!!!!!!

1 answer:

lidiya [134]3 years ago

5 0

Answer:

f(-2) = 0

f(1) = 12

f(2) = 5

Step-by-step explanation:

f(-2) will lie in the function $f(x) =x^2 - 4$

$\therefore f( - 2) = {( - 2)}^{2} - 4 = 4 - 4 = 0$

f(1) will lie in the function $f(x) =2x^2 +10$

$\therefore f(1) = 2 {(1)}^{2} + 10 = 2 + 10 = 12$

f(2) will lie in the function $f(x) =5$

$\therefore f(2) = 5$

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Find the distance between E(-3,4) and B(-1,6)

Marrrta [24]

Answer:

The answer is

<h2> $2 \sqrt{2 } \: \: or \: \: 2.828 \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3> $d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$ </h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

E(-3,4) and B(-1,6)

The distance between them is

<h3> $|EB| = \sqrt{ ({ - 3 + 1})^{2} + ({4 - 6})^{2} } \\ = \sqrt{ ({ - 2})^{2} + ({ - 2})^{2} } \\ = \sqrt{4 + 4} \ \ \\ = \sqrt{8}$ </h3>

We have the final answer as

<h3> $2 \sqrt{2 } \: \: or \: \: 2.828 \: \: \: units$ </h3>

Hope this helps you

4 0

3 years ago

A container contains 12 diesel engines. The company chooses 5 engines at random, and will not ship the container if any of the

Rufina [12.5K]

Answer: $\frac{^{10}C_5}{^{12}C_5}$

Step-by-step explanation:

Given a total of 12 diesel engines

out of which 2 are defective

Company have to choose 5 good engines to ship.

Therefore no of ways in good engines are selected is $^{10}C_5$

no of ways in which any 5 engines are selected from a total of 12 engines is

$^{12}C_5$

Therefore the required probability = $\frac{^{10}C_5}{^{12}C_5}$

= $\frac{7}{22}$ =0.318

3 0

3 years ago

In a sample of 41 temperature readings taken from the freezer of a restaurant, the mean is 29.7 degrees and the population stand

yKpoI14uk [10]

Answer: $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

Step-by-step explanation:

Given: Sample size: n = 41

Sample mean $\overline{x}= 29.7$ degrees

Population standard deviation $\sigma=2.7$ degrees

Confidence level (c) = 80% =0.80

Significance level (a) = 1- c = 1-0.80 = 0.20

z-score for 80% confidence level : z = 1.2816 [from z-table]

Confidence level for population mean :-

$\overline{x}\pm z\dfrac{\sigma}{\sqrt{n}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{\sqrt{41}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{6.403124}$

$=29.7\pm ( 1.2816)(0.42167)$

$=29.7\pm 0.5404$

$=(29.7-0.5404,\ 29.7+0.5404)$

$=(29.1596,\ 30.2404)$

Hence, 80% confidence interval for the temperatures in the freezer $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

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