What is the domain of the function?

To the nearest tenth, what is the area of the shaded segment when AG = 6 ft?

valentinak56 [21]

A. 10.3 assuming that 113.1 is given

7 0

3 years ago

Read 2 more answers

Heeeeeeeeelp!!!!!!!!!!!!!!!!

alekssr [168]

Hello,

(√11-√3)(√11+√3)(√11-√5)=(11-3)(√11-√5))=8(√11-√5)
=8√11-8√5

6 0

3 years ago

Three roots of the polynomial equation x4 - 4x3 - 2x2 + 12x + 9 = 0 are 3, -1 and -1. Explain why the fourth root must be a real

LiRa [457]

Descares Theorem says that the number of roots of a polynomial is equal to its highest exponent, and the roots could be real and/or imaginary (complex).

However is their is one complex root (a + i.b), it's conjugate is also
a root (a-i.b).

In the equation x⁴ - 4x³ - 2x² + 12x + 9 = 0, we have already 3 roots, then the 4rth cannot be a complex one since we don't have a fifth for its conjugate.

3 0

3 years ago

I Need Help Guys , Been in it for 30 minutes

kodGreya [7K]

Answer:

The ratio is 2:3. 10m:15m is the fraction 10/15. Reduce to 2/3. The ratio is 2:3.

Step-by-step explanation:

4 0

2 years ago

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)

skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Step-by-step explanation:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample x⁻ = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

Step 2:-

The 95% of confidence intervals for the average spending

( $(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

$(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )$

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{28-30 }{\frac{10}{\sqrt{25} } }$

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

Conclusion:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0

3 years ago