What us 2 1/3 divided by 1 1/2 in mixed number form?

Jill is factoring the expression 13XY -52y. Her work is shown below

Sunny_sXe [5.5K]

Answer:13y(x-4)

Step-by-step explanation:

13xy-52y =13y(x-4)

3 0

3 years ago

Do anyone know the answer to this? Answers please?

Amanda [17]

Answer:

Yes

Step-by-step explanation:

You can check whether the ratios form a proportion, by setting one piece of the ratios as x

Thus, we can use the equation 4/3.2=22/x.

From this, we get 4x=70.4.

Solving for x gives us 17.6.

5 0

2 years ago

A sales manager gathered information on the number of sales calls made and the number of copiers sold for a random sample of 20

lubasha [3.4K]

Answer:

. The sales manager gathered information on the numbers of sales calls made and the number of copiers sold for a random sample of sales representative. Is there a positive correlation between calls made and copiers? Test at the 0.05 level of significance. Determine the 90% prediction interval for 60 number of calls made. * Calls, X Sold. Y 20 40 20 50 40 60 50 90 40 80 20 40 40 60 30 60

5 0

2 years ago

Pleaassseeeee helpppppp,,,,!!!!!!

Flura [38]

Surface Area (SA) of a Cylinder formula:
SA = 2*pi*r*h+2*pi*r^2
If you fill in the formula:
SA = 2*3.14*2*9+2*3.14*2^2
Simplify:
SA = 6.28*18+6.28*4
SA = 113.04 + 25.12
SA = 138.16 in
I hope I helped
If its wrong I'm super duper extra sorry!

6 0

3 years ago

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past

Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

Let $\mu$ = average miles for deluxe tires

So, Null Hypothesis, $H_0$ : $\mu \geq$ 50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, $H_A$ : $\mu$ < 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is One-sample z test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean lifespan = 46,500 miles

$\sigma$ = population standard deviation = 8000 miles

n = sample of tires = 28

So, test statistics = $\frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }$

= -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

4 0

3 years ago