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3 years ago

What is the solution to……………………ASAP

Mathematics

1 answer:

leonid [27]3 years ago

5 0

Answer:

first option

Step-by-step explanation:

Using the rules of logarithms

log x - log y = log ( $\frac{x}{y}$ )

log $x^{n}$ ⇔ n log x

ln e = 1

Given

4 + 5 $e^{x+2}$ = 11 ( subtract 4 from both sides )

5 $e^{x+2}$ = 7 ( divide both sides by 5 )

$e^{x+2}$ = $\frac{7}{5}$ ( take ln of both sides )

ln $e^{x+2}$ = ln ( $\frac{7}{5}$ )

(x + 2) lne = ln ( $\frac{7}{5}$ )

x + 2 = ln ( $\frac{7}{5}$ ) ( subtract 2 from both sides )

x = ln( $\frac{7}{5}$ ) - 2

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If a drink costs 2.32, how many drinks can you buy with 13 dollars?

Llana [10]

Answer:

5 drinks

Step-by-step explanation:

If the drinks cost $2.32 and you have $13 you can divide 13 by 2.32

13/2.32+ 5.6034...

You cannot buy a partal drink so you look at only the whole number which is 5.

6 0

3 years ago

Read 2 more answers

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

$P(\= X \ge 51 ) =0.0062$

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

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