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3 years ago

What is the shape of a cross section formed by a slice that is parallel to the base of a triangular pyramid?

Mathematics

1 answer:

Shalnov [3]3 years ago

8 0

Triangle because is one pyramid

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-0.75 is it rational or irrational

Mademuasel [1]

-0.75 is a rational number because it can be expressed as the fraction -3/4 and it is an integer

6 0

3 years ago

Read 2 more answers

If the price of rail season tickets increases by 4% every year, work out the price after 3

Zarrin [17]

Answer:

3048 x 1.04 ^3 = 3428.585474

4 0

3 years ago

5 1/2+ 3 2/3+ 1 4/5

Mazyrski [523]

Answer:

$10\frac{29}{30}$

Step-by-step explanation:

$5\frac{1}{2} +3\frac{2}{3} +1\frac{4}{5}$

Let us group the integers and the fractions.

$5\frac{1}{2} +3\frac{2}{3} +1\frac{4}{5}$ = $(5+3+1)+(\frac{1}{2} +\frac{2}{3}+\frac{4}{5} )$

= $9+\frac{15+20+24}{30}$

$= 9+\frac{59}{30}$

$=9+1\frac{29}{30}$

$=10\frac{29}{30}$

7 0

3 years ago

In 2015 there were 759 laptops at lightning high school starting a 2016 the school box 75 more laptops at the end of each year t

nikdorinn [45]

Answer:

Step-by-step explanation:

Given the following :

Number of laptops in 2015 = 759

t(x) = 75x + 759 ; model to determine number of laptops at the school. X years after 2015.

Number of laptops in the school in 2020:

4 0

3 years ago

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

1 year ago