A water heater tanks holds 150 gallons. 3% of the water is lost as steam. how many gallons remain to be used?

7 log(15/16)+6log(8/3)+5log(2/3)+log(32/25) by logarithm laws

hammer [34]

Answer:

1.5871

Step-by-step explanation:

According to the Question,

we have, 7 log(15/16)+6log(8/3)+5log(2/3)+log(32/25)

The Solution, With Basic Properties of Log.

7*log15-7*log16+6*log8-6*log3+5*log2-5*log3+log32-log25

7*1.176-7*1.204+6*0.903-6*0.477+5*0.3010-5*0.477+1.505-1.3979

on Solving, We get

1.5871

4 0

3 years ago

You were given the polynomial f(x)=3x^3+13x^2+19x+k and you know that f(-2)=0find k

eduard

Answer:

k = 10

Explanation:

The initial polynomial is:

$f(x)=3x^3+13x^2+19x+k$

If f(-2) = 0, then when we replace x by -2, the result will be 0. It means that we can write the following equation:

$\begin{gathered} f(-2)=3(-2)^3+13(-2)^2+19(-2)+k=0 \\ 3(-2)^3+13(-2)^2+19(-2)+k=0 \end{gathered}$

Therefore, we can solve for k as follows:

$\begin{gathered} 3(-8)+13(4)+19(-2)+k=0 \\ -24+52-38+k=0 \\ -10+k=0 \\ -10+k+10=0+10 \\ k=10 \end{gathered}$

So, the value of k is 10

6 0

1 year ago

3b-4/2 = c (solve for b) please show work, thanks!

liq [111]

Answer:

Step-by-step explanation:

1. multiply both sides by 2

3b-4=2c

2. add 4 to both sides

3b=2c+4

3. divide both sides by 3

b= $\frac{2b+4}{3}$

6 0

4 years ago

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 12.1 fluid ounc

Ludmilka [50]

Answer:

$\mu = 12.1\\\sigma = 0.3$

a) Find the probability that the drink is less than 11.9 fluid ounces

We are supposed to find P(x<11.9)

x= 11.9

We will use z score formula :

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

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

Refer the z table

So, P(z<-0.67) = 0.2514

Hence the probability that the drink is less than 11.9 fluid ounces is 0.2514

b)Find the probability that the drink is between 11.6 and 11.9 fluid ounces

x= 11.6

We will use z score formula :

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

$z=\frac{11.6-12.1}{0.3}$

$z=-1.67$

x= 11.9

We will use z score formula :

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

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

We are supposed to find P(11.6<x<11.9)

So, P(11.6<x<11.9)= P(x<11.9)-P(x<11.6)

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

Refer the z table

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

P(-1.67<z<-0.67)= 0.2514-0.0475

P(-1.67<z<-0.67)= 0.2039

So, the probability that the drink is between 11.6 and 11.9 fluid ounces is 0.2039.

c)Find the probability that the drink is more than 12.6 fluid ounces

We are supposed to find P(x>12.6)

x= 12.6

We will use z score formula :

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

$z=\frac{12.6-12.1}{0.3}$

$z=1.67$

Refer the z table

P(x>12.6)=1-P(x<12.6)

P(z>1.67)=1-P(z<1.67)

P(z>1.67)=1-0.9525

P(z>1.67)=0.0475

Hence the probability that the drink is more than 12.6 fluid ounces is 0.0475

d) Is the drink containing more than 12.6 fluid ounces an unusual event?

Since p value is less than alpha (0.05)

Answer = No, because the probability that a drink containing more than 12.6 fluid ounces is less than 0.05, this event is not unusual.

3 0

3 years ago

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of $114.00. Assume the population standard deviation is $15.30. Construct a 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, $\bar x$ = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

$CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}$

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

$CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}$

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have $CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}$

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

8 0

3 years ago

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