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Ronch [10]
3 years ago
7

Expand the equation (p+q)^5

Mathematics
1 answer:
VashaNatasha [74]3 years ago
7 0
(p + q)⁵
(p + q)(p + q)(p + q)(p + q)(p + q)
{[p(p + q) + q(p + q)][p(p + q) + q(p + q)](p + q)}
{[p(p) + p(q) + q(p) + q(q)][p(p) + p(q) + q(p) + q(q)](p + q)}
(p² + pq + pq + q²)(p² + pq + pq + q²)(p + q)
(p² + 2pq + q²)(p² + 2pq + q²)(p + q)
{[p²(p² + 2pq + q²) + 2pq(p² + 2pq + q²) + q²(p² + 2pq + q²)](p + q)}
{[p²(p²) + p²(2pq) + p²(q²) + 2pq(p²) + 2pq(2pq) + 2pq(q²) + q²(p²) + q²(2pq) + q²(q²)](p + q)}
(p⁴ + 2p³q + p²q² + 2p³q + 4p²q² + 2pq³ + p²q² + 2pq³ + q⁴)(p + q)
(p⁴ + 2p³q + 2p³q + p²q² + 4p²q² + p²q² + 2pq³ + 2pq³ + q⁴)(p + q)
(p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴)(p + q)
p⁴(p + q) + 4p³q(p + q) + 6p²q²(p + q) + 4pq³(p + q) + q⁴(p + q)
p⁴(p)+ p⁴(q) + 4p³q(p) + 4p³q(q) + 6p²q²(p) + 6p²q²(q) + 4pq³(p) + 4pq³(q) + q⁴(p) + q⁴(q)
p⁵ + p⁴q + 4p⁴q + 4p³q² + 6p³q² + 6p²q³ + 4p²q³ + 4pq⁴ + pq⁴ + q⁵
p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵
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SOMEONE PLEASE HELP!<br>Set up an inequality showing that the radicand cannot be negative.
Stolb23 [73]

Answer:

3t-9 >= 0

Step-by-step explanation:

Original function f(t)= square root 3t-9

There cannot be a negative in the radicand therefore 3t-9 cannot equal a negative number.  

6 0
3 years ago
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In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older
Viktor [21]

Answer:

p_v =P(t_{(63)}>2.5)=0.0075  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=45 represent the mean height for the sample  

s=16 represent the sample standard deviation for the sample  

n=64 sample size  

\mu_o =40 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean age is higher than 40 years, the system of hypothesis would be:  

Null hypothesis:\mu \leq 40  

Alternative hypothesis:\mu > 40  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{45-40}{\frac{16}{\sqrt{64}}}=2.5    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=64-1=63  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(63)}>2.5)=0.0075  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

6 0
3 years ago
-6d=-42 can you help me solve please
eduard

Given the equation :

-6d=-42

So, to find d , divide both sides by ( -6 )

So,

\begin{gathered} \frac{-6d}{-6}=\frac{-42}{-6} \\  \\ d=7 \end{gathered}

Note: when we divide two negative numbers, the result will be a positive number

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Which is an arithmetic sequence?
puteri [66]
B) is an arithmetic sequence because you are adding by 2 each time. The other sequences don’t match this criteria.
I’m sorry I’m not sure what else to say I hope this somewhat helped :/
Have a great day :)
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Is y=6x-2 proportional
Alexus [3.1K]
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