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Mandarinka [93]

3 years ago

9

A company makes wax candles shaped like rectangular prisms. Each candle is 5 cm long, 4 cm wide, and 15 cm tall. If the company

used 5100 cm of wax, how many candles did they make?

2 answers:

faust18 [17]3 years ago

8 0

Answer:

idk do you know this is a hard one

Step-by-step explanation:

inessss [21]3 years ago

7 0

Answer:

5100 cm of wax will make 17 candles

Step-by-step explanation:

Each candle is 5 cm long, 4 cm wide, and 15 cm tall

to know the space they cover, we divide it with their volume and since they are in rectangular prism

Volume is length x width x height

5 x 4 x 15 = 20 x 15 = 300cm

1 candle = 300cm

what number of candle is represented as x

1 candle comprises of 300cm wax

x = 5100

300x = 5100 x 1 =5100

300x= 5100

x= 5100/300 = 17

5100 cm of wax will make 17 candles

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A random sample of 384 people in a mid-sized city (city one) revealed 112 individuals who worked at more than one job. A second

Svetlanka [38]

Answer:

99% confidence interval is:

(0.00278 < P1 - P2< 0.15921)

Step-by-step explanation:

For calculating a confidence intervale for the difference between the proportions of workers in the two cities, we calculate the following:

$[(p_{1} - p_{2}) \pm z_{\alpha/2} \sqrt{\frac{p_{1}(1-p_{1})}{n_{1}} + \frac{p_{2}(1-p_{2})}{n_{2}} }$

Where $p_{1}$ : proportion sample of individuals who worked

at more than one job in the city one

$n_{1}$ : Number of respondents in the city one

$p_{1}$ : proportion sample of individuals who worked

at more than one job in the city two

$n_{1}$ : Number of respondents in the city two

Then

α = 0.01 and α/2 = 0.005

and $z_{\alpha/2} = 2.575$

$p_{1} = \frac{112}{384} = 0.2916$

$p_{2} = \frac{91}{432} = 0.2106$

$n_{1}= 384$ and $n_{2}= 432$

The confidence interval is:

$[(0.2916 - 0.2106) \pm 2.575 \sqrt{\frac{0.2916(1-0.2916)}{384} + \frac{0.2106(1-0.2106)}{432} }$

(0.00278 < P1 - P2< 0.15921)

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3 years ago

The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where paramet

never [62]

Answer:

a) $t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$

The degrees of freedom are given by:

$df=n-1=45-1=44$

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

b) $p_v =P(t_{(44)}>2.744)=0.0044$

Step-by-step explanation:

Information given

$\bar X=18.4$ represent the sample mean

$s=2.2$ represent the sample standard deviation

$n=45$ sample size

$\mu_o =17.5$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test if the true mean is higher than 17.5, the system of hypothesis would be:

Null hypothesis: $\mu \leq 17.5$

Alternative hypothesis: $\mu > 17.5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing we got:

$t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$

The degrees of freedom are given by:

$df=n-1=45-1=44$

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

Part b

The p value would be given by:

$p_v =P(t_{(44)}>2.744)=0.0044$

5 0

4 years ago

How to write a paragraph proof? Given: ∠T and ∠V are right angles TW║VU.<br> Prove: ΔTUW≅ΔVWU

mafiozo [28]

Note: The missing diagram is attached below for a reference.

Answer:

Δ $TUW$ ≅ Δ $VWU$ by $AAS$ case.

Step-by-step explanation:

Given:

∠T and ∠V are right angles.

$TW$ ║ $VU$

To Prove:

Δ $TUW$ ≅ Δ $VWU$

As

∠ $T$ ≅ ∠ $V$ ≅ $90^{0}$ (Given)

Side $UW$ ≅

$TW$ ║ $UV$

and $UW$ is a transverse

So,

∠ $TWU$ ≅ ∠ $WUV$ $\left[alternate\:\:interior\:\:angles\right]$

$As\:Angle\:=\:Angle\:=\:side\:are\:equal$

Thus,

Δ $TUW$ ≅ Δ $VWU$

Keywords: proof, paragraph, logic, reason

Learn more about proof and reason from brainly.com/question/13396866

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3 years ago

The length of a particular rectangle is 4 greater than the width of the rectangle. If the perimeter of the rectangle is 16, what

Nuetrik [128]

2l+2w=perimeter <= equation for perimeter
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<span>2w+2w+8=24 <= solve the equation </span>
<span>4w=16 </span>
<span>w=4 </span>
<span>and, since </span>
<span>w+4=l <= length is 4 greater than width </span>
<span>then l=4+4 </span>
<span>l=8</span>

4 0

4 years ago

Read 2 more answers

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Sergeu [11.5K]

Answer:

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Step-by-step explanation:

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3 years ago