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2 years ago

Can someone please show me how 19169000*e^(0.15)= 222712201 When e= 2.7182818284?

Mathematics

1 answer:

Ierofanga [76]2 years ago

8 0

Answer:

It doesn't. x = 22 271 201

Step-by-step explanation:

You are going to need a calculator no matter how you do it.

$x =19 169 000 \times e^{0.15}$

(a) The direct method

$x = 19 169 000\times 2.718 281 828^{0.15} = 19 169 000 \times 1.161 834 243 = 22 271 201$

(b) The indirect method

$\ln \left (19 169 000\times 2.718 281 828^{0.15} \right ) = \ln(19 169 000) + 0.15 = 16.768 805 + 0.15 = 16.918 804\\\\e^{16.918 804} = 22 271 201$

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One can convert temperature from Kelvin into Fahrenheit using the formula F=95(K−273)+32. What is the temperature in Kelvin corr

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Answer:

The temperature in Kelvin corresponding to F degrees Fahrenheit is $K=\frac{F-32}{95}+273$

Step-by-step explanation:

Given : One can convert temperature from Kelvin into Fahrenheit using the formula $F=95(K-273)+32$ .

To find : What is the temperature in Kelvin corresponding to F degrees Fahrenheit?

Solution :

Temperature from Kelvin into Fahrenheit is given by,

$F=95(K-273)+32$

Writing Kelvin in terms of Fahrenheit,

Subtract 32 both side,

$F-32=95(K-273)$

Divide by 95 both side,

$\frac{F-32}{95}=K-273$

Add 273 both side,

$\frac{F-32}{95}+273=K$

Therefore, The temperature in Kelvin corresponding to F degrees Fahrenheit is $K=\frac{F-32}{95}+273$

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Given f(x) = 7x^9, find f^-1 (x). Then state whether f^-1 (x) is a function

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A though without graphing software it would appear to be C

Step-by-step explanation:
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3 years ago

The weight of National Football League (NFL) players has increased steadily, gaining up to 1.5 lb. per year since 1942. Accordin

GuDViN [60]

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable X can be defined as the weight of National Football League (NFL) players now.

The mean weight is, μ = 252.8 lb.

The standard deviation of the weights is, σ = 25 lb.

A random sample of n = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The sample of players selected is quite large, i.e. n = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

$\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)$

Compute the probability that the sample mean weight will be more than 262 lb as follows:

$P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z$

*Use a z-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.

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