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2 years ago

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Andrew is x years old. Blake is 7 years younger than Andrew. in 5 years time Blake will be twice the age of Carla. how old Carla

now in terms of x?

1 answer:

Lady_Fox [76]2 years ago

4 0

Answer: Carla is 1/2x years old

Step-by-step explanation:

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1 point) the solution of a certain differential equation is of the form y(t)=aexp(3t)+bexp(4t), y ( t ) = a exp ⁡ ( 3 t ) + b ex

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<span>y(t) = a exp(3t) + b exp(4t) conditions, y(0) = 3 y'(0) = 3 y(0) = a exp(3 x 0) + b exp(4 x 0) = a exp(0) + b exp(0) = (a x 1) + (b x 1) = a + b y'(0) = 0 so the linear equation is, a + b = 3</span>

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2 years ago

Been a while since I’ve needed to not use a calculator. Why is -2X10^-18 (3/16) = 4.09X10^-19?

Anestetic [448]

Answer:

see the explanation

Step-by-step explanation:

we have

$2.18*10^{-18}(\frac{3}{16})$

we have that

$\frac{3}{16}=0.1875$

Convert to scientific notation

$0.1875=1.875*10^{-1}$

substitute

$2.18*10^{-18}(1.875*10^{-1}})$

Remember that

To multiply two numbers in scientific notation, multiply their coefficients and add their exponents

$2.18*10^{-18}1.875*10^{-1}}=(2.18*1.875)*10^{(-18-1)}=4.0875*10^{-19}$

Round the number 4.0875 two decimal places

so

$4.0875*10^{-19}=4.09*10^{-19}$

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2 years ago

Describe the transformations needed to translate the graph y=1/x to the graph of y=1/x+3 -4

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i cant describe but i put y=1/x+3 -4 on the graph hope this helps

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2 years ago

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You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

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2 years ago

Click this link to view O'NET's Skills section for Municipal Clerks Note that common skills are listed toward the top

Alika [10]

Writing, reading comprehension

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2 years ago

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