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Keith_Richards [23]
4 years ago
7

What is wrong with the equation? π 4 sec(θ) tan(θ) dθ = 4 sec(θ) π π/3 = −12 π/3 There is nothing wrong with the equation. f(θ)

= 4 sec(θ) tan(θ) is not continuous on the interval [π/3, π] so FTC2 cannot be applied. f(θ) = 4 tan(θ) is not continuous on the interval [π/3, π] so FTC2 cannot be applied. f(θ) = 4 sec(θ) is not continuous at θ = π/3 so FTC2 cannot be applied. The lower limit is not equal to 0, so FTC2 cannot be applied.
Mathematics
1 answer:
satela [25.4K]4 years ago
5 0

Answer:

if f(θ) = 4 is true then print"hello" elif print "ues'

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Prior to a special advertising campaign, 23% of all adults recognized a particular companyâs logo. At the close of the campaign
pochemuha

Answer:

The answer is "2.4049"

Step-by-step explanation:

Calculating the test of Hypothesis: H_{0}: 23\% \ \text{off all adults which reconize the compony's logo}\\\\H_{1}: \text{more than 23\% of adult recornise the compony's logo}\\\\

that is

H_{0}: p=0.23\ against \ H_{1}:p>0.01\\\\Z=\frac{P-p}{\sqrt{\frac{p(1-p)}{n}}}\sim N(0,1)\\\\

Given:

p= 0.23\\\\ \therefore \\\\1-p=0.77\\\\n=1200\\\\ P=\frac{311}{1200}=0.2591\\\\\therefore\\\\Z= \frac{0.2591-0.23}{\sqrt{((0.23)\times \frac{(1-0.23))}{1200}}}=2.4049

Z=2.576 tabled value. Because Z is 2.4049, that's less than Z stated, there is no indication that a null hypothesis is rejectable, which means that 23% of all adults record the logo of the Company.

8 0
3 years ago
Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to
Novosadov [1.4K]

Answer:

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

X \sim N(\mu = 150000, \sigma =2300

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

7 0
3 years ago
Which of the Roy is equivalent to 18- square root of -25?
deff fn [24]

Answer:

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Step-by-step explanation:

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igomit [66]
Mmm me too, and a big glass of coca cola Yummy! :)
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3 years ago
Read 2 more answers
Please help!!!!!!!!!
Natali [406]

Answer: $432

Step-by-step explanation: If he marks up the price by 80% then you multiply the price (240) times 0.8, which is 192. That is how much he marks it up -- $192. SO then you add how much he marks it up to the original price of $242 which is $432.

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