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valentina_108 [34]

3 years ago

9

Two spinners have been spun 100 times. Using the experimental probabilities below, what is the best guess for the number of time

s these spinners would land on the same number in 10 spins? a) 1, b) 2,c) 7,d) 8 P(A)=21 P(A')=79

1 answer:

Monica [59]3 years ago

4 0

Im gonna say c idk if its right but its worth a shot

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In triangle NQL, point S is the centroid, NS = (x + 10) feet, and SR = (x + 3) feet.

olya-2409 [2.1K]

Where R is the median between Q and L:

From my understanding of a triangle's centroid, it divides an angle bisector into parts of 2/3 and 1/3. In the given problem, these divisions are NS and SR. Therefore, twice SR would be equal to NS. From here, we can get the value of X, to solve for SR.

NS = 2SR
(x + 10) = 2(x + 3)
x + 10 = 2x + 6
x = 4

Therefore, SR = (x + 3) = 7

4 0

3 years ago

Read 2 more answers

Find the matrix product, if possible. Show your work please

steposvetlana [31]

$\left[\begin{array}{ccc}-1&3\\5&4\end{array}\right] \cdot \left[\begin{array}{ccc}0&-2&6\\1&-3&2\end{array}\right]\\\\= \left[\begin{array}{ccc}(-1)(0)+(3)(1)&(-1)(-2)+(3)(-3)&(-1)(6)+(3)(2)\\(5)(0)+(4)(1)&(5)(-2)+(4)(-3)&(5)(6)+(4)(2)\end{array}\right]\\\\= \left[\begin{array}{ccc}0+3&2-9&-6+6\\0+4&-10-12&30+8\end{array}\right]\\\\ =\left[\begin{array}{ccc}3&-7&0\\4&-22&38\end{array}\right]$

3 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

A2 + b2 = n2 because of the

Nataly_w [17]

Phytagorean Theorem i think

3 0

3 years ago

Read 2 more answers

One angle of an isosceles triangle measures 138°. Which other angles could be in that isosceles triangle?

BartSMP [9]

Answer:

2 21 degree angles

Step-by-step explanation:

An isosoles triangle has two congruent angles and every triangle's sum of interior angles equals 180 degrees. 138*2>180, so the 138 degree angle cannot be congruent to any of the other angles. Therefore:

180=138+2x

42=2x

21 degrees=x

3 0

3 years ago