Find the surface area of the square pyramid using its net

Every rational number is: (a) an integer (b) a real number (c) a natural number (b) a whole number

mixer [17]

Answer:

Option C

A Natural Number

Step-by-step explanation:

Every rational number is natural number

-TheUnknownScientist

3 0

3 years ago

Read 2 more answers

Look at pic for question and answer choices.

Doss [256]

The answer would b the 3rd option

7 0

3 years ago

How much money should be deposited today in an account that earns 7% compounded semiannually so that it will accumulate to $11,0

Alona [7]

Answer:

$8950.37

Step-by-step explanation:

Use the compound amount formula A = P(1 + r/n)^(nt), in which P is the initial amount of money (the principal), r is the interest rate as a decimal fraction, n is the number of times per year that interest is compounded, and t is the number of years.

Here we have A = $11,000, n = 2, r = 0.07 and t = 3, and so:

$11,000 = P(1 + 0.07/2)^(2*3), or

$11,000 = P (1.035)^6

$11,000 $11,000

Solving for P, we get P = ---------------- = ------------- = $8950.37

1.035^6 1.229

Depositing $8950.37 with terms as follows will result in an accumulation of $11,000 after 3 years.

7 0

3 years ago

What is 3/4 - 9/12. plz help

vova2212 [387]

3 divided by 4 subtract 9 divided by 12 equals 0. Hope this helps

8 0

3 years ago

Read 2 more answers

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago