Bob gets paid 64 hour heard that is 2/3 Melissa makes how much is Alicia paid per hour

Use the digits 1-7 to crest a seven digit number that can be rounded off to 6,300,000

Ne4ueva [31]

6,277,765 rounds up to 6,300,00

7 0

2 years ago

HELLLPPP

Travka [436]

4+4+4+6+6=24 Hours
15+15+20+20+20= 1 Hr 30 Min
24+1= 25 Hours + 30 Minutes=
Final answer: C) 25 hours and 30 minutes

4 0

2 years ago

Help me plz thanks my friend

NikAS [45]

Answer:

I believe the answer would be 21.

Step-by-step explanation:

the shadow of the tall flag is three times the small flag, so the height of the flag should also be three times the height. which is 21

6 0

2 years ago

How do you write 19,830 in scientific notation? _____× 10^_____

pshichka [43]

Answer:

1.983(10⁴)

Step-by-step explanation:

Since we need to have ones and decimals as proper scientific notation, we have 1.983. Since we need the value of 19830, we need to move the decimal place 4 places to the right, so our exponent is 4.

*I double checked my work this time.

8 0

2 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.

7 0

2 years ago