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3 years ago

7

To estimate the slope of the tangent line at t = 15, we average the slopes of the adjacent secant lines for t = 10 and t = 20. A

s obtained in part (a), those slopes are msec = −148.8 and msec = −112.8, respectively. Therefore, the slope of the tangent line at t = 15 is as follows. (In the last step, round your answer to one decimal place.)

1 answer:

Nata [24]3 years ago

4 0

Answer:

-130.8

Step-by-step explanation:

The average of the two slope values is ...

(-148.8 -112.8)/2 = -261.6/2 = -130.8

The slope of the tangent line at t=15 is approximately -130.8.

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at the start of 2014 lucys house was worth 200000. the value of the house is increased by 5% every year. work out the value of h

Galina-37 [17]

Answer:

231525

Step-by-step explanation:

2014 = 200,000

Start of 2015

200,000 x 1.05 = 210,000

Start of 2016

210 x 1.05 = 220,500

Start of 2017

220,000 x 1.05 = 231,525

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3 years ago

In a race in which twelve automobiles are entered and there are no ties, in how many ways can the first three finishers come i

k0ka [10]

That is 3 slots
12 in 1st slot
11 in 2nd
10 in 3rd

so that is 12*11*10 or 1320 ways

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3 years ago

Read 2 more answers

If a rectangle is 2 meters more than its width, and the area of it is 80 square meters, what is the length and width?

bezimeni [28]

Ok I think it means that the length $l$ of the rectangle is 2 m more than is width $w$ ...
The area is A= $l*w=(w+2)*w=80$
so: $w^{2} +2w-80=0$
Solving this equation (using the quadratic formula) you get 2 solutons. The only acceptable is the positive one, i.e. $w=8 m$ , the length will then be $l=w+2=8+2=10m$ .

6 0

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

Can someone please help me with numbers 1, a, b, c, 2, a, b, c

algol [13]

Thank you thank you very much

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3 years ago