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marta [7]
3 years ago
7

Sam and Bruno were computing how many kilometers they rode in the 3 bike trips they took last month. In order, they rode 45.7, 4

0.9, and 38 miles. Their solutions are shown.
If one kilometer equals 0.621 miles, which is the best solution to find the total number of kilometers they rode? Why is it correct?
Bruno is correct. When going from a larger unit to a smaller unit you need to divide.
Bruno is correct. When going from a smaller unit to a larger unit you need to divide.
Sam is correct. When going from a larger unit to a smaller unit you need to multiply.
Sam is correct. When going from a smaller unit to a larger unit you need to multiply.
Mathematics
2 answers:
djverab [1.8K]3 years ago
4 0

Answer:

in easier form bruno is correct

Step-by-step explanation:

Ivenika [448]3 years ago
3 0

We are given number of miles the 3 bike trips they took last month = 45.7, 40.9, and 38 miles.

Sam and Bruno were computing how many kilometers.

Also one kilometer equals 0.621 miles.

From this we can see than 1 kilometer is smaller unit and a mile is a larger unit.

In order to convert number of miles into kilometers, we need to divide each number of mile by 0.621 miles.

We always divide a larger value by a smaller value to get the smaller unit.

Therefore,

<h3>Bruno is correct. When going from a larger unit to a smaller unit you need to divide.</h3>
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You deposit $7,900 in a money-market account that pays an annual interest rate of 4.3%. The interest is compounded quarterly. Ho
aivan3 [116]
A=7,900×(1+0.043÷4)^(4×3)
A=8,981.57
5 0
3 years ago
Deon has a pizza with a diameter
OverLord2011 [107]

The square box is enough to fit the pizza with a diameter of 10 inches inside. Since the area of the square box is more than the area of the pizza, the pizza fits easily in the square box.

<h3>What is the area of the circle and the square?</h3>

The area of the circle is

Ac = πr² = πd²/4 sq. units

Where r is the radius and d is the diameter of the circle.

The area of the square is given by

As = s² sq. units

Where s is the length of the side of a square.

<h3>Calculation:</h3>

It is given that a pizza(in a circular shape) with a diameter d = 10 in is to be placed in a square box of the same length as the diameter of the pizza.

So,

The area of pizza is

Ap = Ac = πd²/4 sq. units

     = π(10)²/4

     = 25π

     = 78.54 sq. in

Then, the area of the square box with the length same as the diameter of the pizza is,

As = d²

    = 10²

    = 100 sq. in

Since the area of the square is more than the area of the pizza (100 sq. inch > 78.54 sq. inch), the pizza easily fits into the square box.

Learn more about the area of a circle here:

brainly.com/question/15673093

#SPJ1

7 0
2 years ago
Sean bought 8 slurpees at $1.50 each and 6 popsicles. He gave the cashier $25.00 and received $4.00 change. How much did each Po
Lina20 [59]

Answer:

1.50

Step-by-step explanation:

8 x 1.50=12

25-12=13

13-4=9

9/6=1.5

8 0
3 years ago
Simplified (4x3 - 5x2 + 3x) + (-2x3 - x2 + 6x)
bogdanovich [222]

Answer:

<em>Answer is</em><em> </em><em>imaginary</em><em> </em><em>root</em><em>s</em>

Step-by-step explanation:

({4x}^{3}  -  {5x}^{2}  + 3x) + ( -  {2x}^{3}  -  {x}^{2}  + 6x) \\ 2 {x}^{3}  - 6 {x}^{2}  + 9x = 0 \\ dividing \: by \: x \\ 2  {x}^{2}  - 6x + 9 = 0 \\ dividin g\: by \: 2 \\  {x}^{2}  - 3x +  \frac{9}{2}  = 0 \\ shifting \: the \: constant \: term \: on \: right \: side \\  {x}^{2}  - 3x =  -  \frac{9}{2}  \\ adding \:  \frac{1}{2}  \times (coefficient \: of \: x) \: whole \: square \\

On solving the above mentioned equation we get some imaginary values.

<em> </em><em> </em><em> </em><em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

6 0
3 years ago
An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e
uranmaximum [27]

Answer:

<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

7 0
3 years ago
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