An equation written below. 3/4x -4 = -19 what is the solution to the equation?

Based on the table, which best predicts the end behavior of the graph of f(x)?

Yuki888 [10]

Answer: A. As x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞.

6 0

3 years ago

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Round your answer to this problem to the nearest degree.

Vika [28.1K]

Use the sine law: sinB/b = sinA/a

SinB = b*sinA/a

sinB = 3*sin120° /8

∠ B = 18.94° ≈ 19°

5 0

3 years ago

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Solve equation by completing the square 5n^2-3n-10=15

VARVARA [1.3K]

These are the two answers you could get

5 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

The low temperatures for 5 days was 5, -2, -11 ,0 and-3.What was the average low temperature for those days

Daniel [21]

Data:
Average (A): ?
Arithmetic mean (A.M) = 5, -2, -11, 0, -3sum of n numbers = 5

Solving:
$A = \frac{AM}{n} = \frac{5+(-2)+(-11)+0+(-3)}{5} = \frac{5-2-11-3}{5} = \frac{\diagup\!\!\!\!3-11-\diagup\!\!\!\!3}{5} = \frac{-11}{5} = - 2.2$

Answer:

<span>The average low temperature of these days is -2.2</span>

3 0

3 years ago