In 2016, London elected its first _______

HELP ME! Look at the stem-and-leaf plot. What is the mode of the numbers?

S_A_V [24]

Given:

The stem-and-leaf plot or a date set.

To find:

The mode of data set.

Solution:

From the given stem-and-leaf plot, we get the numbers of the data set.

64, 67, 70, 70, 71, 75, 76, 78, 78, 80, 82, 82, 88, 91, 93, 94, 97, 98, 100, 100, 100

Mode of a date set is the most frequent value.

In the given data set the most frequent value is 100 with frequency 3.

Therefore, the mode of data set is 100. Hence, option A is correct.

7 0

3 years ago

What percent of the ninth grade students scored between 78 and 96?

kiruha [24]

Answer:

30%

Step-by-step explanation:

3 0

3 years ago

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Fill in the blank for Remainder Therom<br><br><br> x 3 + x 2 - 21x - 45 = (x + 3)(x - 5)(x+?)

icang [17]

Not sure about remainder theorem but I'm sure that the last terms should all multiply tho the last term

see the expanded form is -45

so the last terms of each binomial should multiply to -45

3 times -5 times ?=-45
-15 times ?=-45
divide by -15
?=3

the question mark is 3

5 0

3 years ago

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What is these decimal numbers from least to greatest

e-lub [12.9K]

Least to Greatest:
1.037, 1.073, 1.37, 1.703

7 0

3 years ago

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Mathematical induction, prove the following two statements are true

adelina 88 [10]

Prove:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}$
____________________________________________

Base Step: For n=1:
$n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1$
and
$4-\dfrac{n+2}{2^{n-1}}=4-3=1$
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}$
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}$

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
$=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}$

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
$1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}$

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
$=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}$

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
$=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}$

Distribute the -2 and combine the fractions together,
$=4+\dfrac{-2k-4+(k+1)}{2^{k}}$

Combine like-terms,
$=4+\dfrac{-k-3}{2^{k}}$

pull the negative back out,
$=4-\dfrac{k+3}{2^{k}}$

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
$=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}$

3 0

3 years ago

In 2016, London elected its first ________ mayor.