All categories

3 years ago

The interquartile range (IQR) for thedata:32,7,3,-5,1,8, 10,36 is :

Mathematics

1 answer:

GenaCL600 [577]3 years ago

7 0

Answer:

Your Interquartile range (IQR) would be 19.

Step-by-step explanation:

-5, 1, 3, 7, 8, 10, 32, 36

Median: 7.5

Lower quartile: 2

Upper quartile: 21

Interquartile range: 21 - 2 = 19

You might be interested in

Mrs. Rogers sold 49 out of 78 crafts at the fair. What percent of her crafts did she sell? Show your

igor_vitrenko [27]

Answer:

62.8, 63% if you around it off.

Step-by-step explanation: Do 49 Divided by 78, then times it by 100 to get your percentage. I just rounded it off to the nearest tenth and whole number just in case. Hope this helped!

3 0

2 years ago

Mark can make 424242 birthday cakes in 777 days.

Zolol [24]

Mark can make 303,030 birthday cakes.

8 0

3 years ago

Read 2 more answers

Convert 27day 8hour into second

Nataliya [291]

27 days and 8 hours = 656 hours
27 x 24 = 648
648 + 8 = 656
24 hours = 3,600 seconds
656 x 3,600 = 2,361,600 seconds

The answer is 2,361,600 seconds.

7 0

2 years ago

A pan of warm water (46∘ Celsius) was put in a refrigerator. After 15 minutes, the water's temperature was 27∘ C; 15 minutes aft

Cloud [144]

Answer:

The refrigerator was at around 13 ∘C

Step-by-step explanation:

Newton's Law of Cooling:

The rate of change of a body temperature (amount of heat loss/time of loss) is directly proportional to the difference between its own temperature and the surroundings.

$\frac{T(t2)-T(t1)}{t2-t1} = -h (T(t2) - Tenv)$

T ⇒ temperature

t ⇒ time

Tenv ⇒refrigerator temperature

$\frac{T(t2)-T(t1)}{t2-t1}$ ⇒ rate of change of he temperature

-h ⇒ constant of proportionality (negative because the temperature is decreasing inside the refrigerator)

We have 3 points:

time (minutes) - Temperature (∘ C)

0 (when the pan was put in the refrigerator) - 46

15 (after 15 minutes) - 27

30 (15 minutes after the first 15 minutes) - 19

$\frac{27 - 46}{15 - 0}$ = -h (27 - Tenv)

$\frac{19 - 27}{15 - 0}$ = -h (19 - Tenv)

Now we have a system of two equations and two variables

$\frac{-19}{15} = -h (27 - Tenv)\\ \frac{19}{15} = h (27 - Tenv)\\ \frac{19}{15(27 - Tenv)} = h$

$\frac{19 - 27}{15 - 0} = -h (19 - Tenv)\\\frac{-8}{15} = -h (19 - Tenv)\\\frac{8}{15} = h (19 - Tenv)\\\frac{8}{15} = \frac{19}{15(27 - Tenv)} (19 - Tenv)\\8(27 - Tenv) = 19 (19 - Tenv)\\216 - 8 Tenv = 361 - 19 Tenv\\19 Tenv - 8 Tenv = 361 - 216\\11 Tenv = 145 \\Tenv = 145/ 11\\Tenv = 13. 18$

The refrigerator was at around 13 ∘C

6 0

2 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago

The interquartile range (IQR) for thedata:32,7,3,-5,1,8, 10,36 is :​

The interquartile range (IQR) for thedata:32,7,3,-5,1,8, 10,36 is :