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Mashcka [7]
2 years ago
10

Solve for x 10 * 90⁰ 26

Mathematics
1 answer:
musickatia [10]2 years ago
4 0

Answer:

x=24

Step-by-step explanation:

x=√(26)²-(10)²

x=√676-100

x=√576

x=24

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A vegetable garden with an area of 200 square feet is to be fertilized. If the length of the garden is 1 foot less than three ti
charle [14.2K]
Ok so I like to go in steps with these questions- first draw a picture and identify your variables.

W=width
L= 3w-1

Now we know that length times width gets us area so we plug in our variables into the area equation.

200 = w(3w-1)

When you foil that equation you end up with a quadratic : 3w^2-w-200 = 0

Either factor that or use the quadratic formula to get
w= 8.33 and w= -8

Since you can't have a negative dimension you need to use 8.33 and plug it back into your length equation.

Final answer:

w= 8.33ft
l= 23.99ft

*Now I simplified the decimals a little bit so you end up with 199.8ft^2 for the area so just add a few decimals on here and there*
5 0
3 years ago
Read 2 more answers
Factoring Quadratic Trinomials 6n^2 - n - 40 =
Sophie [7]

Answer:

(3n - 8)(2n + 5)

Step-by-step explanation:

6n² - n - 40

Consider the factors of the product of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term

product = 6 × - 40 = - 240 and sum = - 1

The factors are - 16 and + 15

Use these factors to split the n- term

6n² - 16n + 15n - 40 ( factor the first/second and third/fourth terms )

2n(3n - 8) + 5(3n - 8) ← factor out (3n - 8) from each term

(3n - 8)(2n + 5)

Then

6n² - n - 40 = (3n - 8)(2n + 5)

6 0
3 years ago
What is the value of the expression (−2 4/5)÷(−1.4)
AysviL [449]

Answer:

2

Step-by-step explanation:

3 0
3 years ago
If x=5, evaluate x^5 * x^-3
Artyom0805 [142]

Answer: 25

Step-by-step explanation:

Plug in x to the expression.

5^5 * 5^{-3}\\

When there is the same base of exponents is being multiplied the exponents add together and the base stays the same.

So this simplifies to, 5^2.

This then solves to 25.

3 0
3 years ago
In one area, monthly incomes of technology-related workers have a standard deviation of $650. It is believed that the standard d
Virty [35]

Answer:

There is sufficient statistical evidence to prove that the standard deviation of the technology-related workers and the standard deviation of the non-technology workers are equal.

Step-by-step explanation:

Here we have our null hypothesis as H₀: σ² = s²

Our alternative hypothesis is then Hₐ: σ² ≠ s²

We therefore have a two tailed test

To test the hypothesis of difference in standard deviation which is the Chi squared test given as follows

\chi ^{2} = \dfrac{\left (n-1  \right )s^{2}}{\sigma ^{2}}

Where:

n = Size of sample

s² = Variance of sample = 950²

σ² = Variance of population = 650²

Degrees of freedom = n - 1 = 71 - 1 = 70

α = Significance level = 0.1

Therefore, we use 1 - 0.1 = 0.9

From the Chi-square table, we have the critical value as

1 - α/2 = 51.739,  

α/2 = 90.531

Plugging the values in the above Chi squared test equation, we have;

\chi ^{2} = \dfrac{\left (23-1  \right )950^{2}}{650 ^{2}} = 49.994

Therefore, since the test value within the critical region, we do not reject the null hypothesis, hence there is sufficient statistical evidence to prove that the standard deviation of the technology-related workers and the standard deviation of the non-technology workers are equal.

6 0
3 years ago
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