ROMAN NUMERAL HI TOMMROW

Which addition sentences show you how to find 16-9?

Valentin [98]

This dosent make sense and where are the sentences ;-;

5 0

4 years ago

Divide 1 hour in the ratio 1 : 9 : 10

olganol [36]

Answer:

3 minutes; 27 minutes; 30 minutes

Step-by-step explanation:

Let the common multiplier of the given ratios be x. So, the ratios 1: 9 : 10 will become x, 9x and 10x respectively.

Therefore,

$x + 9x + 10x = 1 \: hour \\ \\ 20x = 60 \: minutes \\ \\ x = \frac{60 \: minutes}{3} \\ \\ x = 3 \: minutes \\ \\ 9x = 9 \times 3 = 27 \: minutes \\ \\ 10x = 10 \times 3 = 30 \: minutes$

Thus, 1 hour in the ratio 1 : 9 : 10 can be divided as:

3 minutes; 27 minutes; 30 minutes

5 0

3 years ago

Angle CMW and angle WML together form angle CML. Angle CMW is a right angle. What must be true about angle CML?

ANTONII [103]

CML must be an obtuse angle.

Since CMW and WML together form CML, and CMW is a right angle, then CML must be larger than a right angle, which makes it obtuse.

3 0

3 years ago

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student. How much would it cost 13

MakcuM [25]

Answer: it would cost a total of $194

Step-by-step explanation:

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student.

If student ticket costs $x, adult ticket would cost x + 2

This means that the cost for an adult is

7 + 2 = $9

The cost of 13 adult tickets to the movie theatre would be

13 × 9 = $117.

The cost of 11 student tickets to the movie theatre would be

11 × 7 = $77

Total cost of 13 adult tickets and 11 student tickets would be

117 + 77 = $194

3 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