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ArbitrLikvidat [17]
4 years ago
7

Given: Rectangle JKLM has an area of 36 square centimeters. Side is at least 4 centimeters long. Prove: KL 9 centimeters Assum

e that a. ____. Then the area of rectangle JKLM is greater than b. _____ , which contradicts the given information that c. _____. So the assumption must be false. Therefore, d. _____.
Mathematics
1 answer:
goldenfox [79]4 years ago
4 0
The answer is shown below:

Rectangle JKLM has an area of 36 square centimeters. The side is at least 4 centimeters long. Prove: KL is 9 centimeters. Assume that a. "KL > 0". Then the area of rectangle JKLM is greater than b.) "40 centimeters² ", which contradicts the given information that c. the area is 40 centimeters². So the assumption must be false. Therefore, d.)  "KL <= 10". 
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A basket contains 11 pieces of fruit: 4 apples, 5 oranges, and 2 bananas. Jonas takes a piece of fruit at random from the basket
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3 years ago
Find the unknown length in the triangle. 15cm and 18cm
vova2212 [387]
-------------------------------------------------------------------------------------
Formula
-------------------------------------------------------------------------------------
a² + b² = c²

-------------------------------------------------------------------------------------
Find the missing leg.
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15² + b² = 18²
b² = 18² - 15²
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Answer: 9.95 cm
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5 0
4 years ago
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
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