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geniusboy [140]
3 years ago
10

The photo is attached

Mathematics
1 answer:
aleksley [76]3 years ago
7 0

D

using the double - angle identity

cos (2A) = cos² A - sin² A = 2cos² A - 1 = 1 - 2sin² A

the right side = 1 - 2sin² (112.5° ) with A = 22.5°

hence 2A = 2 × 22.5° = 45°

thus cos 45° = 1 - 2sin² ( 22.5°)


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A rectangle has vertices located at (–4, 4), (2, 4), (2, –4) and (–4, –4). What is the length of a diagonal of the rectangle?
Allisa [31]

Answer:

A diagonal of this rectangle has length 10.

Step-by-step explanation:

The vertices (-4, 4) and (-4, -4) have the same x-coordinate (-4) and different y-coordinates (4 and -4).  These two points are the endpoints of a vertical side of the rectangle which has length 4 - (-4) = 8.

Similarly, the vertices (2, -4) and (-4, -4) have the same y-coordinate (-4) but different x-coordinates (2 and -4).  To find the horizontal dimension of the rectangle, we calculate 2 - (-4), which comes out to 6.

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5 0
2 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

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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

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