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3 years ago

HELP ASAP!! the graph is in the picture above to reference it!!

Mathematics

2 answers:

Nikitich [7]3 years ago

8 0

I believe it is B and C. (Side note, I got the same computer from K-12 lol I’m in 8th grade and this is my first year, will it get easier?!?)

jenyasd209 [6]3 years ago

5 0

Answer: There are two correct statements, which are:

b. the intersection (1,60) indicates the time in hours when celia and tayvin are the same distance from the city.

c. Celia drove at a faster rate than Tayvin.

Explanation:

The graph shows two linear functions with these characteristics and informations:

dependent variable (vertical axis): distance traveled in miles)

independent variable (horizontal axis): time (hours):

Tayvin's function:

- inicial distance = y-intercept: (0,10) = 10 miles

- speed = slope = rise / run = [ 60 miles - 50 miles ] / [ 1 hour] =

speed = 50 miles / hour

Celia's function:

- initial distance = y-intercept: (0,0) = 0

- speed = slope = rise / run = [60 miles - 0 miles] / [ 1 hour] =

= speed = 60 miles / hour

With that you can go over each answer choice to asses their validity:

a. Tayvin drove a faster rate than Celia.

FALSE. It was shown that Tayvin's speed was 50 miles/hour while Celia's speed was 60 miles/hour, hence Celia was 10 miles/hour faster than Tayvin.

b. the intersection (1,60) indicates the time in hours when Celia and Tayvin are the same distance from the city.

TRUE. The two lines intersect each other at the point with coordinates (1,60), where 1 is the time in hours and 60 is the distance in miles, meaning that both were at the same distance (60 miles) at the same time (1 hour).

c. Celia drove at a faster rate than Tayvin.

TRUE. As stated before, Celia drove 10 miles/hour faster than Tayvin

d. the intersection (1,60) indicates the time in hours when celia and tayvin are traveling at the same speed.

FALSE. As stated in the point b., the intersection point indicates the time in hours when Celia and Tayvin are the same distance from the city; the speed were different all the time, since both kept thier respective uniform speeds (constant slope means constant speed).

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Gabriela has an apartment that costs 30% of her monthly budget, which is $3,575. How much is her rent?

enot [183]

Answer:

Her rent is $1072.5

Step-by-step explanation:

Monthly budget: $3,575

Her rent is 30% of her monthly budget. Since 30% = 0.3, we can multiply our monthly budget with 0.3 to get 30% of it.

30% * 3575 = 0.3 * 3575 = 1072.5

Answer: $1072.5

5 0

3 years ago

Brent claims that more students at his school prefer hot dogs over hamburgers. His friend Shaun doesn’t agree. To settle their a

photoshop1234 [79]

Answer:

Using conditional probabilities it can be shown that the results are influenced by the gender.

Explanation:

To prove that the results are influenced by <em>gender</em> you can calculate both the probability of preferring hot dogs and the conditional probability of preferring a hot dog given that is a female.

If the two results are different the probability of preferring hot dog is dependent on whether the person is a female or a male.

The probability of preferring hot dogs given that is a female is stated by the problem: 34.2%.

The probability of preferring hot dogs by the whole sample is:

Number of males that prefer hot dogs: 184 (stated by the problem)
Number of females that prefer hot dogs:

100% - 34.2% = 65.8%

65.8% of 635 = 0.658 × 635 = 417.83 ≈ 418

Samples size: 542 males + 635 females = 1177

Probability of preferring hot dogs =

number of students that preffer hot dogs / number of students =

(184 + 418) / 1177 = 602 / 1177 = 0.5115 ≈ 51.2%

Thus, the probability of preferring hot dogs given that the student is a female (34.2%) is different from the probability of preferring hot dog for the whole sample, making the results dependent of the gender.

3 0

4 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

(20 POINTS)

Xelga [282]

88 square feet and by the way your only giving 15 points step your game up

4 0

3 years ago

Which equation of a line of best-fit reflects a negative correlation?

Wittaler [7]

You are correct. you are looking for a negative slope. y=mx+b where m is the slope.
1/70=+
-83/-27 = 83/27=+
8=+
-6=-

Let me know if you have questions.

5 0

4 years ago

Read 2 more answers