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2 years ago

Identify the range of the function shown in the graph.

Mathematics

1 answer:

MAVERICK [17]2 years ago

3 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the plots and graphs.

Since, the x axis varies from -infinite to +infinite,

here, the y totally varies from -5 to +5,

hence, the option B.) is correct.

===>

$- 5 \leqslant y \leqslant 5$

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A 180 pound person is hanging onto a 15-foot rope. Assuming the rope is uniform and that it weighs 2 pound per foot, set up (but

aleksklad [387]

Answer:

The answer is "g= 32 FPS"

Step-by-step explanation:

Given value:

$mass = 180\\\\length= 15 \\\\\lambda = \frac{ mass}{ length} = 2 \ \text{pound per foot} \\\\\bold{Formula}\\\\ \to w \int_{a}^{b} F \cdot d \ dx = \int_{0}^{15} (2x+180) \ g \ dx$

by solving the value we get the value of g that is equal to 32.

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2 years ago

Larry mows lawns for $50 an hour. Last week, he gave all his customers a 20% discount off his hourly rate. Mrs. Grinnell hired L

joja [24]

The answer would be D: $108

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3 years ago

Read 2 more answers

uppose that we have collected information on how much a sample of households spend on clothing per year. If we are 90% confident

Verizon [17]

Answer:

the chance that the population mean will either be less than $1,500 or above $2,100 is 10%

Step-by-step explanation:

We are 90% confident that the true population mean willl be between $1,500 and $2,100

This means that there is a 90% probability that the true population mean will be between $1,500 and $2,100.

So 100-90 = 10% probability that it is outside of this interval, that is, either be less than $1,500 or above $2,100.

So the correct answer is:

the chance that the population mean will either be less than $1,500 or above $2,100 is 10%

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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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2 years ago

A plane flies x mph. How far can it go in y hours?

viva [34]

X*y = distance

So the answer would be B.) xy miles

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3 years ago