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Nimfa-mama [501]

3 years ago

10

If 1/3 of students in a class like school and 1/3 of those that like school, like science, what is the fraction of students that

like science?

2 answers:

Amiraneli [1.4K]3 years ago

5 0

1/3 multiplied by 1/3, which is 1/9
:)

victus00 [196]3 years ago

5 0

1/3 x 1/3 = 1/9

1/9 of students in the class like science

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Solving systems of equations with substitution:<br> y= 2x-4 <br>y= -x+5<br><br>help please!

Furkat [3]

<span>y= 2x-4
y= -x+5
so

</span><span>2x-4 = -x+5
2x +x = 5 +4
3x = 9
x =3

y = 2x-4
y =2(3) - 4
y = 6-4
y = 2

answer: (3,2)</span>

4 0

3 years ago

He normal pulse rate of a 13-year-old is between 70 and 100 beats per minute. As part of a health program, the medical staff at

SVEN [57.7K]

From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.

Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10

</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The least value is 76 and the highest value is 97. The lower and the upper quartiles are 85 and 94 respectively while the median is 89.

Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9

The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5

Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>

6 0

3 years ago

Read 2 more answers

Line equations from point/slope

babunello [35]

Answer:

The answer is

$y = \frac{1}{4} x + 3$

Step-by-step explanation:

Plug in and solve the equation. We are trying to find the y-intercept:

$y= \frac{1}{4}x + b \\ 2 = \frac{1}{4} ( - 4) + b \\ 2 = - 1 + b \\ \frac{ + 1 = + 1 \: \: \: \: \: \: \: \: }{3 = b}$

We found the y-intercept, so now we plug it in it's right full spot in the formula.

$y = \frac{1}{4} x + 3$

8 0

3 years ago

HELP ME PLeaSEEEEEEEEEE<br><br><br><br> thank you <3

Tcecarenko [31]

Answer:

2nd option

Step-by-step explanation:

Graph both equations and find the point where they both intersect. In this case, it's (4,2).

3 0

2 years ago

The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =

pentagon [3]

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

$v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du$

so that

$v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du$

$\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}$

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is $s(0)=0\,\mathrm m$ . Then its displacement over the time interval [0, 3] is

$s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}$

c) The total distance traveled is the integral of the absolute value of the velocity function:

$s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt$

$v(t)$ for $0\le t$ and $v(t)\ge0$ for $\cos^{-1}\left(-\frac29\right)\le t\le3$ , so we split the integral into two as

$\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt$

$=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt$

$\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}$

4 0

3 years ago