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3 years ago

Please please explain how to do this I cant and its not good-x^2+4=10

Mathematics

1 answer:

Marta_Voda [28]3 years ago

4 0

So you follow PEMDAS

So we start with the exponent which is -X^2 so it would turn into X

Now we have X+4=10

We bring the four over and we get X=6

Hope this was helpful and ask questions if you still need help:)

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Find the sum of the first 20 terms of the arithmetic sequence 4, -4, -12, -20

Vsevolod [243]

Answer:

The sum of the first 20 terms is -1440.

Step-by-step explanation:

We want to find the sum of the first 20 terms of the arithmetic sequence:

4, -4, -12, -20...

The sum of an arithmetic sequence is given by:

$\displaystyle S=\frac{k}{2}(a+x_k)$

Where k is the number of terms, a is the initial term, and x_k is the last term.

Since we want to find the sum of the first 20 terms, k = 20.

Our initial term a is 4.

Our last term is also the 20th term as we want the sum of the first 20 terms.

To find the 20th term, we can write an explicit formula for our sequence. The explicit formula for an arithmetic sequence is given by:

$x_n=a+d(n-1)$

Where a is the initial term, d is the common difference, and n is the nth term.

Our initial term is 4. From the sequence, we can see that our common difference is -8 since each subsequent term is eight less than the previous term. Therefore:

$x_n=4-8(n-1)$

Then the last or 20th term is:

$x_{20}=4-8(20-1)=4-8(19)=-148$

Therefore, the sum of the first 20 terms are:

$\displaystyle\begin{aligned} S_{20}&=\frac{(20)}{2}\left((4)+(-148))\\&=10(-144) \\&= -1440\end{aligned}$

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2 years ago

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Which strategy can be used to multiply 6 x 198 mentally?

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2 years ago

Customers arrivals at a checkout counter in a department store per hour have a Poisson distribution with parameter λ = 7. Calcul

IgorLugansk [536]

Answer:

(1)14.9% (2) 2.96% (3) 97.04%

Step-by-step explanation:

Formula for Poisson distribution: $P(k) = \frac{\lambda^ke^{-k}}{k!}$ where k is a number of guests coming in at a particular hour period.

(1) We can substitute k = 7 and $\lambda = 7$ into the formula:

$P(k=7) = \frac{7^7e^{-7}}{7!}$

$P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%$

(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours

$P(k\leq2) = P(k=0)+P(k=1)+P(k=2)$

$P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}$

$P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}$

$P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%$

(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:

$P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%$