All categories

3 years ago

Given that 32 = 9, write this in logarithm form.

Mathematics

1 answer:

Gwar [14]3 years ago

6 0

It is

log 39=2

Explanation:

By definition the logarithmic form

log bx=y is equivalent to the exponential form

by=x

Hence from the given 32=9 you have b=3 , y=2 , x=9

hence the logarithmic form is

log 3 9=2

You might be interested in

Can anyone help me with this please

tigry1 [53]

It’s b here is proof of what i got

4 0

3 years ago

Jason consumes 1500 calories and needs to

Lubov Fominskaja [6]

Answer:

100*150=15000 calories

Step-by-step explanation:

100 times 150 is 15000 because you are multipling 2 people

I don't know if it is correct

just check and if it isn't then check some one elses work.

6 0

3 years ago

Sabra joins a cooking club for $8.50 and pays $6.25 each month. She also joins a movie club for $12 and pays $3.75 each month. W

podryga [215]

Simplified expression for total amount spent is 10(2.05 + m)

Step-by-step explanation:

Step 1: Write expression for expense on joining cooking club for m months.

Expense = 8.50 + 6.25 × m

Step 2: Write expression for expense on joining movie club for m months.

Expense = 12 + 3.75 × m

Step 3: Calculate the total amount spent on both clubs

Total Amount = 8.50 + 6.25m + 12 + 3.75m = 20.5 + 10m = 10(2.05 + m)

6 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