-×+1>-2 and 6(2x-3)>-6

The weather forecast warns of morning fog today. you may only be able to see 0.6 mile in front of you. how many feet is this?

Kamila [148]

This a problem of conversion of units. to convert units, you must know at what values are each units equal. for this problem we must convert miles to feet. so 1 mile is equivalent to 5280 ft. so the answer is :
length in feet = 0.6 miles ( 5280 ft / 1 mile)
length in feet = 3168 feet

4 0

3 years ago

Which is an equation of the line that goes through the points ( 4, 1) and

Aloiza [94]

Answer:

none of the above

Step-by-step explanation:

You can try the points in the equations (none works in any equation), or you can plot the points and lines (see attached). You will not find any of the offered answer choices goes through the given points.

___

You can start with the 2-point form of the equation of a line. For points (x1, y1) and (x2, y2) that equation is ...

y = (y2 -y1)/(x2 -x1)·(x -x1) +y1

Filling in the given points, we get ...

y = (3 -1)/(2 -4)·(x -4) +1

y = 2/(-2)(x -4) +1 . . . . . simplify a bit

y = -x +4 +1 . . . . . . . . . simplify more

y = -x +5 . . . . . . . . . . . slope-intercept form

5 0

2 years ago

Write an inequality to represent the situation. A cheese pizza costs at least $5.

Nady [450]

<
- 5 dollars, the V thing sits on top of the line and 5 comes after signifying that it could be 5 dollars or more

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

F(x) = 2x3 + 732 – 4x - 5 g(x) = 3x – 2 Find ( f - g)(x).

Nadya [2.5K]

Answer:

(f-g) (x)=731.............

6 0

2 years ago