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3 years ago

Which measure of central tendency is MOST EASILY affected by outliers

1 answer:

5 0

The average can be easily affected by outliers because an outlier can bring the average much higher or lower depending on the value. For example, if this was the data set

0,50,51,52
the average would be 38.25

But without the 0, the average would be 51.<span />

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Martha and Janet each decide to invest $600 into different accounts. Martha's account earns $4.50 every month. Janet's account e

ehidna [41]

Answer:

<u>Janet's interest is $3.90 every month.</u>

Step-by-step explanation:

First, you need to set up a proportion to find Janet's interest:

0.65%/100 = x/600

Step 1: Cross-multiply.

0.65/100/100 = x/600

(0.65/100)*(600)=x*(100)

39/10=100x

Step 2: Flip the equation.

100x=39/10

Step 3: Divide both sides by 100.

100x/100=39/10/100

x=39/1000 (0.039)

Then move the decimal twice to the left to find the dollar amount:

0.039 ⇒ 3.90

Janet's interest is $3.90 every month.

HOPE THIS HELPS!!! <33333

-Silver

5 0

3 years ago

Read 2 more answers

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
$\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}$

6 0

4 years ago

If the sixth term of a sequence is 128 and the common ratio is 2, then what is the first term?

Lady bird [3.3K]

Geometric progresión:
we can calculate any term, using this rule:

an=a₁*r^(n-1)
a₁=the first term
r=the common ratio

a₆=128
r=2

an=a₁ * r^(n-1)
128=a₁ * 2⁶⁻¹
128=a₁ * 2⁵
128=a₁ * 32
a₁=128/32=4

Solution: a₁=4

3 0

3 years ago

What is the answer to – 5m (6m2-8m)

dexar [7]

Answer:

-20m^2

Step-by-step explanation:

-5m (6m x 2 - 8m)

-5m (12m-8m)

-5m x 4m

=-20m^2

7 0

3 years ago

50 is ____________ times as large as 5.

coldgirl [10]

Answer: 10

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers