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2 years ago

Which expression is equivalent to (2g3+4)2?

Mathematics

2 answers:

snow_tiger [21]2 years ago

7 0

Answer:

Step-by-step explanation:

(2g³ + 4)² = (2g³ + 4)(2g³ + 4)

Each term in the second factor is multiplied by each term in the first factor, that is

2g³(2g³ + 4) + 4(2g³ + 4) ← distribute both parenthesis

= 4 $g^{6}$ + 8g³ + 8g³ + 16 ← collect like terms

=4 $g^{6}$ + 16g³ + 16 → B

pickupchik [31]2 years ago

6 0

Answer:

B. 4g6+16g3+16

Step-by-step explanation:

Hope it helps

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frutty [35]

Answer:

9 is answer . . . . . . . ........

5 0

2 years ago

A(n) ___________________ variable is a qualitative variable such that there is no meaningful ordering or ranking of the categori

Licemer1 [7]

Answer: c. Nominal

Step-by-step explanation:

Level of measurement determines the nature of description .

The four levels of measurements scales are

1. Nominal scale

Categorize the data on the basis of the quality such as Gender, color, etc.

2. Ordinal scale

Order attributes according to their ranks. For example : 1 < 2 < 3 .

3. Interval scale

Describe the feature of the difference between any two categories. For example : Fahrenheit scale to measure temperature.

4. Ratio scale

Consist of the features of nominal, ordinal, and interval measures but in includes a "true zero" point.

For example : Age.

∴ A nominal variable is a qualitative variable such that there is no meaningful ordering or ranking of the categories.

Hence, the correct answer is c. Nominal .

5 0

2 years ago

the Student Government wants to sell magazines subscription to make a profit of at least $3,000. They make $1.25 per m magazine

IRINA_888 [86]

The student Government wants to make a profit of at least $3000 from a magazines sales. At least $3000 simply means the profit should be $3000 and above.

There profit is $1.25 per magazine(m) subscription sold.

Their expenses are as follows

advertisement in school = $100

advertisement in community = $150

let

m = number of magazine sold

profit = selling price - cost price

selling price

$1.25\times m=1.25m$

Cost price

$100+150=\text{\$}250$

Therefore,

$\begin{gathered} 1.25m-250\ge3000 \\ \text{divide through by 1.25} \\ \frac{1.25m}{1.25}-\frac{250}{1.25}\ge\frac{3000}{1.25} \\ m-200\ge2400 \end{gathered}$

8 0

1 year ago

Which statements are true?

Alinara [238K]

Collinear means the points are lying on single straight line
So B)with K G R form a single straight line is correct
Coplanar means the points are in the same plane
So D)with H G K N on the same plane is correct

Overall,B) and D) are correct

6 0

3 years ago

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

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

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

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different 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 amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will 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 different from 3%.

3 0

2 years ago