30m^2+27m-20=0. It’s quad

Solve the Quadratic Equation By Completing the Square<br> b^2-15b-54=0

vladimir1956 [14]

$b^{2}-15b-54=0. \newline
b^{2}-15b=54. \newline
b^{2}-15b+(\frac{15}{2})^{2}=54+(\frac{15}{2})^{2}. \newline
(b-\frac{15}{2})^{2}=\frac{441}{4}. \newline
b-\frac{15}{2}=\pm\frac{21}{2}. \newline
b=\pm\frac{21}{2}+\frac{15}{2}. \newline
b_1 = -3, b_2 = 18.$

3 0

3 years ago

Tina and Tracey are putting 70 things from their present lives into a few boxes, and they are going to bury the boxes so they ca

vlada-n [284]

number of full boxes is 2

things left over is 10

6 0

3 years ago

Read 2 more answers

Geno signed up for a text messaging plan. He pays a $5.00 monthly fee and $0.04 per text message.

Marysya12 [62]

Answer:

The correct answer is [C.) 0.04r + 5.00}

Step-by-step explanation:

hope this hlps

7 0

3 years ago

Read 2 more answers

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.

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3 years ago

The number of students at maritas school decreased to 98 percent of last years number. Currently, there are 1,170 students. How

Salsk061 [2.6K]

1055 students were in the school last year

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