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3 years ago

What is the volume of the pyramid

Mathematics

1 answer:

Vlad1618 [11]3 years ago

6 0

Answer:

A=lw+l(w

2)2+h2+w(l

2)2+h2=14·20+14·(20

2)2+202+20·(14

2)2+202≈1016.84192

Step-by-step explanation:

That might be it,Im sorry tried

You might be interested in

the volume of pyramid a is the volume of pyramid b. if the heigh of pyramid b increases to twice that of pyramid a the new volum

Radda [10]

Answer:

12.259-12.25 890654321

4 0

3 years ago

the mean of 30 numbers is 150.it was detected on rechecking that one value of 165 was wrongly copied as 135.find the correct mea

N76 [4]

Answer:

151

Step-by-step explanation:

m=150cm

n=30

new mean =(m*n+165-135)/n

=(150*30+165-135)/30

=(4500+30)/30

=4530/30

=453/3

=151

therefore

correct mean =151

Hope this helps :D

6 0

3 years ago

A university is interested in whether there's a difference between students who live on

ryzh [129]

It is to be noted that the determination of whether or not there is a significant difference between the two groups will be done using a t test.

A t-test is a statistical test that juxtaposes two samples' means. It is used in hypothesis testing, using a null hypothesis that the variance in group means is zero and an alternative hypothesis that the difference is not zero.

<h3>What are the conditions for using this kind of confidence interval? </h3>

The conditions to use the t test are:

The sample must be independent
The mean of the population and variance must be unknown.
The Box plot is attached.

<h3>What are the degrees of freedom (k) for this test using the conservative method?</h3>

The degrees of freedom (k) to be utilized for this text will be derived using the conservative method given below:

df = [(s₁²/n₁) + (s₂²/n²)/[((s₁²/n₁)²/((n₁-1)) + (s₂²/n₂)²/((n₂-1))]

= [(3.0952/15) + (6.4095/15)]² / [((3.0952/15)²/14) + ((6.4095/15)²/14)]

= 24.965

Hence,

df ≈ 24 (if approximated to the floor)

<h3>What are the sample statistics for this test?</h3>

Recall the the standard deviation of the population are unequal and unknown. This thus requires that we utilize the two-sample unpooled t-test.

Here, H₀ is given as;

$t = \frac{\bar{x_{1} -\bar{x_{2}}}}{\sqrt{\frac{s_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \sim t_{df}$

t = [(3.33333 - 4.13333)]/√[(3.0952/15) + (6.4095/15)]

= - 0.8/√0.6337

t = - 1.005

<h3>What is the 95% confidence interval for the difference between the number of classes missed by each group of students?</h3>

The 95% confidence interval is computed using the following formula:

$(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)$

= - 0.8 ± t₀.₀₂₅,₂₄ (√0.6337)

= - 0.8 ± 2.064 (√0.6337)

= -2.4429, 0.8429

<h3>What is the a 90% confidence interval for the difference between the number of classes missed by each group of students?</h3>

To derive the 90% interval, we state:

$(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)$

= - 0.8 ± t₀.₀₅₀,₂₄ (√0.6337)

= - 0.8 ± 0.685 (√0.6337)

= -2.162, 0.562

<h3>Based on the two confidence intervals computed in parts d and e, what is the conclusion about the differences between the means of the two groups?</h3>

From the intervals computed, we must fail to reject H₀

H₀ : μ₁ = μ₂

It is clear from the above intervals computed from that the differences between the mean of both groups is significant. This is because, zero is included on the two intervals.

Learn more about t-test at;
brainly.com/question/6589776
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