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Dafna11 [192]
3 years ago
10

7848 / 4 what Is the quotient

Mathematics
1 answer:
andriy [413]3 years ago
8 0

\text{Hey there!}

\text{Both numbers are divisible to 4, so divide the numerator and denominator from 4}

\frac{7,848\div4}{4\div4}

\text{7,848}\div4=1,962

\text{4}\div4=1

\text{1,962}\div1=1,962

\boxed{\boxed{\text{Answer: 1,962}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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Hi! Since it is asking for the total, you must add!

145+217=?
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Last month, Korey’s Comics had a gross profit of $2,385 with a monthly revenue of $3,465. Korey expects similar sales this month
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Which equation can be used to find the value of x?
Dafna11 [192]

Answer:

C

Step-by-step explanation:

We know every triangle equals 180 so were just going to take 180 - 42 = 138 then were going to do 138 - 67 = 71  we can check our work by doing

42 + 67+ 71 = 180. We also know 67 + 42 = 109.

Lets look at our answers;

A. 109 = 180 + x  When solved it gave me; x = -71. We got a positive 71 so this is incorrect.

B. x + 42 = 67   When solved it gave me;  x = 25, which is incorrect.

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D. 42 + 67 - x = 180 When solved it gave me; x =  - 71

Therefore the answer is C.

6 0
3 years ago
In a metal fabrication​ process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are n
Leto [7]

Answer:

95% Confidence interval: (14.4537 ,15.1463)

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 feet

Sample mean, \bar{x} = 14.8 feet

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, σ = 0.65 feet

Degree of freedom = n - 1 = 15

95% Confidence interval:

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 15 and}~\alpha_{0.05} = \pm 2.1314  

14.8 \pm 2.1314(\dfrac{0.65}{\sqrt{16}} ) \\\\= 14.8 \pm 0.3463 = (14.4537 ,15.1463)  

is the required confidence interval for the true mean length of rods.

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3 years ago
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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:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

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

Basic Power Rule:

  1. f(x) = cxⁿ
  2. 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:

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

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

<u>Step 3: Integrate Pt. 2</u>

  1. [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
  2. [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
  3. [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}}
  4. 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)
  5. 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
3 years ago
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