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A paper clip is 2.35 cm. How many meters long is the clip?

sineoko [7]

The answer is 0.0235

3 0

3 years ago

Integrate the following. ∫<img src="https://tex.z-dn.net/?f=5x%5E%7B4%7D%20dx" id="TexFormula1" title="5x^{4} dx" alt="5x^{4} dx

Vadim26 [7]

Answer:

$\displaystyle D) {x}^{5} + \rm C$

Step-by-step explanation:

we would like to integrate the following Integral:

$\displaystyle \int 5 {x}^{4} \, dx$

well, to get the constant we can consider the following Integration rule:

$\displaystyle \int c{x} ^{n} \, dx = c\int {x}^{n} \, dx$

therefore,

$\displaystyle 5\int {x}^{4} \, dx$

recall exponent integration rule:

$\displaystyle \int {x} ^{n} \, dx = \frac{ {x}^{n + 1} }{n + 1}$

so let,

$n = 4$

Thus integrate:

$\displaystyle = 5\left( \frac{ {x}^{4+ 1} }{4 + 1} \right)$

simplify addition:

$\displaystyle = 5\left( \frac{ {x}^{5} }{5} \right)$

reduce fraction:

$\displaystyle = {x}^{5}$

finally we of course have to add the constant of integration:

$\displaystyle \boxed{ {x}^{5} + \rm C}$

hence,

our answer is D)

7 0

2 years ago

Read 2 more answers

The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among

jeyben [28]

Answer:

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(c) The value of P(at most one) is 0.253.

(d) The event is not unusual.

Step-by-step explanation:

Let X = number of households watching the show.

The probability of the random variable x is, P (X) = p = 0.18.

The sample selected for the survey is of size, n = 14

The random variable X follows a Binomial distribution with parameter n = 14 and p = 0.18.

The probability of a Binomial distribution is computed using the formula:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...$

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

$P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062$

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-0.062\\=0.938$

Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

$={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}+{14\choose 1}(0.18)^{1}(1-0.18)^{14-1}\\=0.062+0.191\\=0.253$

Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.

5 0

3 years ago

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ represent the value that we want to test

$\alpha$ 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 true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

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

Calculate the statistic

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

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

5 0

3 years ago

8 and 9 please Extra 100 points And brainleist

lesantik [10]

Answer:

9 . (4)

10 . 17

Step-by-step explanation:

5 0

3 years ago