A baby elephant weighs 200 pounds at birth. Seven years later, the elephant's weight is 3,000 pounds.

13. Use the Change of Base Formula to evaluate log373. (1p

Dovator [93]

<h3>Question:-</h3>

$\sf log_{3}(73)$

<h3>Solution:-</h3>

$\sf \frac{In(73)}{In(3)}$

$≈ \frac{4.29046}{1.0986}$

$≈3.905$

3 0

3 years ago

1. Write the slope-intercept form of the equation of each line given the slope and y-intercept.

MAXImum [283]

Answer:

1b. y = -x + 5

1a. y = 5x - 3

Step-by-step explanation:

1.

b. y = -x + 5

a. y = 5x - 3

Just write out the Slope-Intercept Formula as it is given to you.

I am joyous to assist you anytime.

6 0

3 years ago

The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The

Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

The general form of a (1 - α)% confidence interval is:

$CI=SS\pm CV\times SD$

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

$CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)$

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

$37\pm 0.1=SS\pm CV\times SD$

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

$P(-1$

The percentage of z-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of t is:

$t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262$

*Use a t-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0

3 years ago

Points M and N are both on ZB with M between Z and N. ZM = 10

maxonik [38]

Answer:

Your answer is 40%

Step-by-step explanation:

It is 20 units between Z and B.

We add 2 more points, and one of them is at the the halfway point of the entire line.(M)

One of them is only 2 units away from B.(N)

That means our line is halved, so from 20/20 possible places a point could go and land in the right area is now 10/20.

Minus 2 from that; to take into count of the N to B, and you have 8/20

8/20 is just 40%

5 0

3 years ago

In the number 1934 is a value of 9 in 100 place 10 times the value of the 3 in the tens place

dybincka [34]

no because mathematically 30X10=300 not 900 its would have to be 30X30 to get the product of 900 so therefore no your statement is incorrect

8 0

3 years ago