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Nata [24]
3 years ago
10

How many pounds.are in 96 ounces

Mathematics
2 answers:
Kryger [21]3 years ago
7 0
Because we know that there are 16 ounces in a pound, we can just divide the number of ounces (96) by 16 to find out the equivalent number of pounds. 96/16 = 6 Therefore, 96 ounces is equal to 6 pounds. Hope this helps! :)
UkoKoshka [18]3 years ago
4 0
16 ounces are in 1 pound. So all you need to do is divide 96 by 16. Your answer is 6 pounds. To check, multiply 6 and 16 to get 96. Hope this helps
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Given that S is the centroid of triangle MNO, find SQ.
AnnZ [28]

Answer:

|SQ|=5

Step-by-step explanation:

If S is the median, then OP is a median of triangle OMN.

This implies that:

|MP|=|NP|

3x-4=x+4

We group like terms and solve for x.

3x-x=4+4

\implies 2x=8

\implies x=4

Now we know that: MN:SQ=2:1

But MN=2x+2

This implies that:

2x+2:SQ=2:1

Put x=4

2(4)+2:SQ=2:1

10:SQ=2:1

Therefore |SQ|=5

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3 years ago
Which value of v is a solution to this equation? 3v=24
zhannawk [14.2K]
V = 24/3 = 8
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6 0
3 years ago
Read 2 more answers
Look at the following pairs of numbers (x, y).
katrin2010 [14]

Step-by-step explanation:

Taking the first coordinate point (3,16.5)

where x= 3 and y= 16.5

\frac{y}{x}  =  \frac{16.5}{3}

\frac{y}{x}  = 5.5

y = 5.5x

optionB

8 0
3 years ago
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A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

<u>Step-by-step explanation</u>:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0
3 years ago
A drawing of the state of Utah is 7 cm tall and 5.4 cm wide. If the scale used to create the drawing was 1 cm to 50 miles, how m
Ksivusya [100]

Answer:

5

Step-by-step explanation:

4 0
3 years ago
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