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2 years ago

What is the lateral area of the pyramid

Mathematics

1 answer:

Mkey [24]2 years ago

3 0

Answer:

43.75 ft²

Step-by-step explanation:

$s_l$ = (l√(w/2)² + h²) + (w√(l/2)² + h²)

l & w become 3.5, and h becomes 6.

$s_l$ = (3.5√(3.5/2)² + 6²) + (3.5√(3.5/2)² + 6²)

Step 1:Because this is a square pyramid, what you see above essentially becomes what you see below.

$s_l$ = 2(3.5√(3.5/2)² + 6²)

Step 2: Divide 3.5 by 2 to get 1.75.

$s_l$ = 2(3.5√1.75² + 6²)

Step 3: Square both 1.75 and 6 to get 3.0625 and 36 respectively.

$s_l$ = 2(3.5√3.0625 + 36)

Step 4: Add 3.0625 and 36 to get 39.0625.

$s_l$ = 2(3.5√39.0625)

Step 5: The square root of 39.0625 is 6.25.

$s_l$ = 2(3.5 * 6.25)

Step 6: Multiply 3.5 by 6.25 to get 21.875.

$s_l$ = 2(21.875)

Step 7: Multiply 2 by 21.875 to get 43.75.

$s_l$ = 43.75 ft²

The lateral area of this pyramid is 43.75 ft².

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Company A makes a large shipment to Company B. Company B can reject the shipment if they can conclude that the proportion of def

boyakko [2]

Answer:

$z=\frac{0.1475-0.1}{\sqrt{\frac{0.1(1-0.1)}{400}}}=3.17$

The p value for this case would be given by:

$p_v =P(z>3.17)=0.00076$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 0.1 and then Company B can reject the shipment

Step-by-step explanation:

Information provided

n=400 represent the random sample taken

X=59 represent number of defectives from the company B

$\hat p=\frac{59}{400}=0.1475$ estimated proportion of defectives from the company B

$p_o=0.1$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true proportion of defectives is higher than 0.1 then the system of hypothesis are.:

Null hypothesis: $p \leq 0.1$

Alternative hypothesis: $p > 0.1$

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.1475-0.1}{\sqrt{\frac{0.1(1-0.1)}{400}}}=3.17$

The p value for this case would be given by:

$p_v =P(z>3.17)=0.00076$

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3 years ago

Use the formula c=d and = to find the diameter of this circle whose circumference equals 172.7cm

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Answer

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2 years ago

Read 2 more answers

Use the information given to answer the question. Vanessa and Moe record their quiz scores in music class for a semester. The do

Lorico [155]

Answer:

Vanessa records a total of 11 quiz scores

Moe records a total of 10 quiz scores

According to the data, Vanessa's mean score is 80

According to the data, Vanessa's median score is 85

The range of Moe's quiz score is 45

The interquartile range of Moe's quiz score is 16.25

Step-by-step explanation:

Vanessa's score is given in the dot plot

Moe's scores are; 75, 80, 90, 90, 95, 55, 100, 95, 85, 85

1) The number of quiz scores Vanessa records is given by the number of dots on the dot plot, n = 11 dots ≡ 11 quiz scores

2) The number of quiz scores Moe records is given by the number of data points, n = 10 data points ≡ 10 quiz scores

3) Vanessa's mean score = ∑x $_i$ /n

∑x $_i$ /n = (60 + 2× 65 + 75 + 4 × 85 + 2 × 90 +95)/11 = 80

Vanessa's mean score = 80

4) Vanessa's median score is the (11 + 1)/2 th score = Her 6th score = 85

Vanessa's median score = 85

5) The range of Moe's score = Her largest score - Her least score = 100 - 55 = 45

The range of Moe's score = 45

6) Sorting Moe's scores gives;

75, 80, 90, 90, 95, 55, 100, 95, 85, 85

55, 75, 80, 85, 85, 90, 90, 95, 95, 100

The first quartile, Q₁ = 11/4th score = 2.75th score = 75 + 0.75 × 5 = 78.75

The 3rd quartile, Q₃ = The 3/4×11 score = 8.25th score = 95 + 0.25 ×0 = 95

The interquartile range, I = Q₃ - Q₁ = 95 - 78.75 = 16.25

The interquartile range of Moe's quiz score = 16.25.

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Step-by-step explanation:

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