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3 years ago

Which of the following points is a solution of y > |x| + 5?

Mathematics

2 answers:

morpeh [17]3 years ago

8 0

Answer:

B. (1,7)

Step-by-step explanation:

Answer is B. (1,7)

If x = 1 then

y > 1 + 5

7 > 6

malfutka [58]3 years ago

7 0

Answer:

(1 , 7) is a solution of y > IxI + 5 ⇒ answer B

Step-by-step explanation:

* Lets revise the absolute value

- IxI = positive value

- IxI can not give negative value

- The value of x could be positive or negative

* Lets solve the problem

∵ y > IxI + 5

∴ y > x + 5 <em>OR</em> y > -x + 5

- Lets check the answers

∵ y > 0 + 5 ⇒ y > 5

- But y = 5, and 5 it is not greater than 5 and there is no difference

between the two cases because zero has no sign

∴ (0 , 5) not a solution

∵ y > 1 + 5 ⇒ y > 6

- Its true y = 7 and 7 is greater than 6

∵ y > -1 + 5 ⇒ y > 4

- Its true y = 7 and 7 is greater than 4

∴ (1 , 7) is a solution

∵ y > 7 + 5 ⇒ y > 12

- But y = 1 and 1 is not greater than 12

∵ y > -7 + 5 ⇒ y > -2

- Its true y = 1 and 1 is greater than -2

* we can not take this point as a solution because it is wrong

with one of the two cases

∴ (7 , 1) is not a solution

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The slope of the line that passes through the points is 0 slope. That is B is the answer.

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Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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