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3 years ago

How many 3-letter combinations can be made from this set if no letter can be repeated?

Mathematics

1 answer:

Wittaler [7]3 years ago

7 0

Answer:

B 120 different combinations

Step-by-step explanation:

There are 6 letters to choose from

Picking a letter for the first spot 6

Then there are 5 letters left

Picking a letter for the second spot 5

Then there are 4 letters left

Picking a letter for the second spot 4

6*5*4

= 120

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Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

zloy xaker [14]

Answer:

$k=0$

$\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}$

$\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}$

Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

$\displaystyle k = \ln 2$

Therefore:

$2k+1 = 2\ln 2+ 1 \approx 2.3863$

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$

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3 years ago

Modern Electronics offers a one year monthly

Marysya12 [62]

Answer:

Step-by-step explanation:

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MrMuchimi

Answer:

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In a one-way ANOVA, what is the difference between the among-groups variance MSA and the within-groups variance MSW?

LuckyWell [14K]

Answer:

$MSR=MSA=\frac{SSR}{k-1}$

And $MSE=MSW=\frac{SSE}{N-k}$

The difference is that MSA takes incount the variation between the groups and the grand mean, and the MSW takes in count the variation within groups respect to the mean of each group .

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have $j$ groups and on each group from $j=1,\dots,j$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}=6750+8000=14750$

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1$ where k represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-k$ .

And the total degrees of freedom would be $df=N-1$

We can find the $MSR=MSA=\frac{SSR}{k-1}$

And $MSE=MSW=\frac{SSE}{N-k}$

The difference is that MSA takes incount the variation between the groups and the grand mean, and the MSW takes in count the variation within groups respect to the mean of each group .

And the we can find the F statistic $F=\frac{MSR}{MSE}=$

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Find the midpoint between two points on a coordinate plane whose coordinates are (5,7) and (-3,1)

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Step-by-step explanation:

Mid point formula ,

x = (x1+x2)/2 , y = (y1+y2)/2

x = 1 , y = 4

therefore ,midpoint is (x, y)= (1,4)

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