a 12 foot tall building casts on 8 foot long long shadow. how long of a shadow will a 5 foot tall women have

A country's population in 1993 was 94 million. in 1999 in was 99 million. estimate the population in 2005 using the exponential

Jet001 [13]

The initial population is
P₀ = 94 million in 1993

The growth formula is
$P(t) = P_{0}e^{kt}$
where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.

Because P(5) = 99 million (in 1999),
$94e^{5k} = 99 \\e^{5k}=1.0532 \\ 5k = ln(1.0532) \\ k = 0.010367$

In the year 2005, t = 12 years, and
$P(12)=94e^{0.010367*12} = 106.45$

Answer: 106 million (nearest million)

7 0

1 year ago

will crown brainliest Stephanie puts 30 cubes in a box. The cubes are 1/2 inch on each side. The box holds two layers with 15 cu

KiRa [710]

15 cubic inches is the answer hope this helps and please give me brainiest

5 0

1 year ago

Read 2 more answers

the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of

soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = $\sigma$ = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

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

Using the values, we get:

$z=\frac{25-22}{4}=0.75$

So,

P(X ≥ 25) is equivalent to P(z ≥ 0.75)

Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

$-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20$

Thus, a test score of 20 represent the 31st percentile of the distribution.

3 0

2 years ago

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

arlik [135]

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

Now we have everything in order to replace into formula (1):

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$

$-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087$

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0

1 year ago

Sandy has 16 roses,8 daisies, and 32 tulips. She wants to arrange all the flowers in bouguets. Each bouguet has the same number

avanturin [10]

Answer:

Greatest number of flowers that can be used in a bouquet is 8.

Step-by-step explanation:

In this question greatest number of flowers used in a bouquet will be decided by the "Greatest Common Factor" of the numbers of the flowers given.

So, factors of 16 = 1×2×2×2×2

Factors of 8 = 1×2×2×2

Factors of 32 = 1×2×2×2×2×2

Common factors = 1×2×2×2

Greatest Common Factor of these numbers will be = 1×2×2×2 = 8.

Therefore, greatest number of flowers that could be in a bouquet is 8.

3 0

1 year ago