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3 years ago

An initial investment of $200 is now valued at $350. The annual interest rate is 8% compounded continuously. The

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

7 years

Step-by-step explanation:

An initial investment of $200 is now valued at $350. The annual interest rate is 8% compounded continuously. The equation represents the situation, where t is the number of years the money has been invested. About how long has the money been invested? Use a calculator and round your answer to the nearest whole number.

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In a market, 44 of the 80 types of vegetables are grown locally. What percent of the vegetables are grown locally? Identify the

VladimirAG [237]

44 is the part.
80 is the whole.

$\frac{44}{80} =\frac{x}{100}$

We need to cross multiply, which is: a/b = c/d ⇒ ad=bc

After cross multiplying, we get:

$44 \times 100 = 80 \times x\\\\4400 = 80x\\\\\sf{Divide ~80~ on~ both~ sides}\\\\\frac{80x}{80} =\frac{4400}{80}\\\\x =\frac{4400}{80} \\\\\boxed{\bf{52.3809524\approx 52.38}}$

Approximately 52.38% of the vegetables are grown locally.

5 0

3 years ago

A semicircle has an area of 76.97 squared km. Find it’s radius

shtirl [24]

Answer:

7 km

Step-by-step explanation:

Substitute:

The equation for area of a semi-circle is $\frac{r^{2}\pi }{2}$

A = $\frac{r^{2}\pi }{2}$

76.97 = $\frac{r^{2}\pi }{2}$

Multiply by 2 on both sides:

2(76.97) = ( $\frac{r^{2}\pi }{2}$ )2

153.94 = $r^{2} \pi$

Divide by pi on both sides:

153.94 = $r^{2} \pi$

/pi /pi

49.03 ≈ $r^{2}$

Find the square root:

$\sqrt{49.03} =\sqrt{r^{2} }$

7 = r

7 0

3 years ago

last week the daily Low temperatures for a city in degrees celsuis were 5 8 6 5 10 7 and 1 what was the average low temperature

Tems11 [23]

Answer:

Step-by-step explanation:

5+8+6+5+10+7+1=42

42/7=6

7 0

3 years ago

Multiply using FOIL or Table method (x + 4) (x + 5)

mylen [45]

Answer:

${x}^{2} + 9x + 20$

Step-by-step explanation:

1) Use FOIL method: (a + b) (c + d) = ac + ad + bc + bd.

${x}^{2} + 5x + 4x + 20$

2) Collect like terms.

${x}^{2} + (5x + 4x) + 20$

3) Simplify.

${x}^{2} + 9x + 20$

Therefor, the answer is x² + 9x + 20.

5 0

3 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago