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3 years ago

How do you graph f(x)=3

Mathematics

1 answer:

Yuri [45]3 years ago

7 0

A horizontal line through 3 on the y axis

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Find the volume of 22 ft

Mazyrski [523]

V=4πr^3V=4π(11)^3<span>
V=5575.28
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4 years ago

Drag the tiles to the correct boxes to complete the pairs. Match each expression with its simplified form. Tiles 3a5 − (-a5) 6a7

fredd [130]

3a^5 - (-a^5) = 3a^5 + a^5 = 4a^5

6a^7 - 2a^7 = 4a^7

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7a^7 - 8a^7 = -a^7

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4 years ago

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Find two integers whose sum is -3 and product is -18

Mariulka [41]

let two integers be x and y

A.T.Q

x+y= -3

or, x= -3-y(i)

and xy= -18

or,(-3-y)y= -18[by(i)]

or,- -3y-y²= -18

or,-y²= -18+3

or, -y²= -15

or, y²=15

therefore y=✓15

from (i)

x= -3-✓15

y=√15

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3 years ago

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HELP ASAP!!! 74 POINTS AND BRAINLIEST!!!!

Mice21 [21]

Answer:

the first one goes in the botten and the 2nd one goes into top 3rd goes into bottem.

Step-by-step explanation:

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3 years ago

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

valkas [14]

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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3 years ago