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2 years ago

It was found that an unknown solid had a density of 20 g/cm3 and a volume of 30 cm3. Determine the mass of the solid.

Mathematics

1 answer:

In-s [12.5K]2 years ago

4 0

Answer:

Step-by-step explanation:

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A pack of cinnamon-scented pencils sells for $8.00. What is the sales tax rate if the total cost of the pencils is $8.64?

vesna_86 [32]

Selling price of a pack of cinnamon scented pencils = $8
Cost of the pack of pencils along with tax = $8.64
Then
Amount of tax taken = (8.64 - 8.00) dollars
   = 0.64 dollars
Percentage of sales tax taken = (0.64/8.00) * 100
   = 64/8
   = 8 percent
So the rate of sales tax for the pack of cinnamon scented pencils is 8%. I hope the procedure is clear enough for you to understand. You can use this method for solving similar problems without requiring any help from outside.

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2 years ago

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Exactly one of the die is a four

Ksju [112]

Care to explain more ??

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2 years ago

What’s the answer for this?

poizon [28]

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3 years ago

a television with a 5:4 screen shows an image with a ratio of 20:12 which creates a letter boxed image. What percent of the scre

Grace [21]

Answer: $75\%$

Step-by-step explanation:

You know that the ratio of the television's screen is $5:4$ . This ratio can be also written as the following fraction:

$\frac{5}{4}$

The television shows an image with a ratio of $20:12$ which can be also written as a fraction:

$\frac{20}{12}$

Set up the following proportion and solve for "x":

$\frac{20}{x}=\frac{5}{4}\\\\x=\frac{4*20}{5}\\\\x=16$

Now, in order to find the percent of the screen area occupied make the following division:

$\% occupied=\frac{12}{16}*100=75\%$

6 0

2 years ago

A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random samp

motikmotik

Answer:

There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.

Step-by-step explanation:

Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.

We find mean = 11.015

Sample std deviation = 3.157

a) $H_0: \bar x= 12 oz\\H_a: \bar x >12$

(Right tailed test)

Mean difference /std error = test statistic

$\frac{11.015-12}{\frac{3.157}{\sqrt{10} } } \\=-0.99$

p value =0.174

Since p >0.01, our alpha, fail to reject H0

Conclusion:

There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.

3 0

2 years ago