What is the Surface Area Formula for a Square Pyramid? I need the formula

Mathematics

1 answer:

Contact [7]4 years ago

8 0

Answer:

The surface area of a square pyramid is the sum of the areas of all its 4 triangular side faces with the base area of the square pyramid. If a, h, and l are the base length, the height of the pyramid, and slant height respectively, then the surface area of the square pyramid = a2+ 2al (or) a2+2a √a24+h2 a 2 4 + h 2 .

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Write the quadratic function with the following transformations: vertical shrink by 3, left 4, down 9.

allsm [11]

Y=x^2
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3 0

4 years ago

Please can someone help me with this

Eduardwww [97]

Answer:

$\boxed{V = 434.9 \ cm^3 }$

Step-by-step explanation:

Volume of Sphere = $\frac{4}{3} \pi r^3$

Where r = 4.7 cm

V = $\frac{4}{3} (3.14)(4.7)^3$

V = $\frac{4}{3} (3.14)(103.8)$

V = $\frac{1304.7}{3}$

V = 434.9 cm³ (Up to 1 dp)

4 0

4 years ago

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A racing car consumes a mean of 100 gallons of gas per race with a variance of 64. If 44 racing cars are randomly selected, what

lesantik [10]

Answer:

83.89% probability that the sample mean would be greater than 98.8 gallons

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$