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3 years ago

Rachel has tossed a fair coin ten times. It landed heads up every time. Is this POSSIBLE?

1 answer:

4 0

Yes, although this is unlikely as every time she flips the coin there is a 50% chance of getting heads. Therefore, there is a possibility of her getting heads every time

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PLEASE HELP ME WITH BOTH MATH QUESTIONS ASAP!!!!!!!!!

maxonik [38]

Answer:

the one on top= c= -12

the bottom one= f= -15

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

3 years ago

Maria takes 2 hours to plant 50 flower bulbs. Lois takes 3 hours to plant 45 flower bulbs. Working together, how long should it

kherson [118]

Answer:

Maria: (50 bulbs)/(2 hours) = 25 bulbs/hour

Lois: (45 bulbs)/(3 hours) = 15 bulbs/hour

Together: 25 + 15 = 40 bulbs/hour

(150 bulbs)/(40 bulbs per hour) = 3 3/4 hours

(3/4 hours)(60 minutes/hour) = 45 minutes

Total time: 3 hours 45 minutes

6 0

2 years ago

Read 2 more answers

Help asap..............

MatroZZZ [7]

The common denominator for 25 and 20 would be 100

25 x 4 = 100

20 x 5 = 100

Rewrite 14/25 as 14 x 4 / 25 x 4 = 56/100

Rewrite 11/20 as 11 x 5 / 20 x 5 = 55/100

14/25 is greater than 11/20

14/25 > 11/20

3 0

3 years ago

How do I simplify (6^-2)(3^-3)(3 x 6)^4? Not the answer, just the simplified expression.

Mazyrski [523]

The simplification form of the provided expression is 108 after applying the integer exponent properties.

<h3>What is integer exponent?</h3>

In mathematics, integer exponents are exponents that should be integers. It may be a positive or negative number. In this situation, the positive integer exponents determine the number of times the base number should be multiplied by itself.

We have an expression:

$=\left(6^{-2}\right)\left(3^{-3}\right)\left(3\:\cdot \:6\right)^4$