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3 years ago

5

Select the correct location on the number line which point on the number line represents the approximate volume of a cylinder wi

th a radius of 4 units and a height of 4 units? Use 3.14 for TT

1 answer:

Vikki [24]3 years ago

6 0

V = 3.14(4^2) × 4
V = 3.14(16) × 4
V = 50.24 × 4
V = 200.96
According to this, answer is E.

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Find the sum of 7 1/4 and 4 2/3

wariber [46]

Answer:

I hope you got it and plz mark brainliest !

5 0

3 years ago

What is 85 into 7000

Marizza181 [45]

Answer:

Step-by-step explanation:

7000/85 = 1400/17

or 82.352941

8 0

2 years ago

The perimeter of a rectangular field is 320320 yd. the length is 2020 yd longer than the width. find the dimensions.

hichkok12 [17]

The sum of length and width is half the perimeter, 160 yd. The length is the average of the sum and difference of dimensions:
.. (160 +20)/2 = 90
Then the width is 160 -90 = 70

The rectangular field is 70 yd by 90 yd.

6 0

2 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

Use the "rule of 72" to estimate the doubling time (in years) for the interest rate, and then calculate it exactly. (Round your

Law Incorporation [45]

Answer:

According to the rule of 72, the doubling time for this interest rate is 8 years.

The exact doubling time of this amount is 8.04 years.

Step-by-step explanation:

Sometimes, the compound interest formula is quite complex to be solved, so the result can be estimated by the rule of 72.

By the rule of 72, we have that the doubling time D is given by:

$D = \frac{72}{Interest Rate}$

The interest rate is in %.

In our exercise, the interest rate is 9%. So, by the rule of 72:

$D = \frac{72}{9} = 8$ .

According to the rule of 72, the doubling time for this interest rate is 8 years.

Exact answer:

The exact answer is going to be found using the compound interest formula.

$A = P(1 + \frac{r}{n})^{nt}$

In which A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

So, for this exercise, we have:

We want to find the doubling time, that is, the time in which the amount is double the initial amount, double the principal.

is double the initial amount, double the principal.

$A = 2P$

$r = 0.09$

The interest is compounded anually, so $n = 1$

$A = P(1 + \frac{r}{n})^{nt}$

$2P = P(1 + \frac{0.09}{1})^{t}$

$2 = (1.09)^{t}$

Now, we apply the following log propriety:

$\log_{a} a^{n} = n$

So:

$\log_{1.09}(1.09)^{t} = \log_{1.09} 2$

$t = 8.04$

The exact doubling time of this amount is 8.04 years.

4 0

3 years ago