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2 years ago

I will give BRAINLIEST!!! middle school math...

Mathematics

2 answers:

Aloiza [94]2 years ago

5 0

Answer:

A. Q(2, -1), S(-6,5)

Step-by-step explanation:

Reflection rule across the x-axis:

(x, y) → (x, -y)
P(2, 1) → Q(2, -1)
R(-6, -5) → S(-6, 5)

Correct choice is A

JulijaS [17]2 years ago

3 0

Answer:

Solution given:

A(x,y))----reflection about x axis--->A'(x,-y)

P(2,1)---reflection about x axis----->Q(2,-1)

R(-6,-5)----reflection about x axis-->S(-6,5)

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A number is multiplied by 6 and the result is 48. Find the number.

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Answer:

the answer is 8

Step-by-step explanation:

8x6=48

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2 years ago

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the art teacher is sending home a total of 3,150 projects before winter break if each of her 150 students had the same number of

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Answer:

21 projects each

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3 years ago

Max points :)) 33 please help no explanation needed..

Xelga [282]

5(2y-4) - 3y = 1. 10y-20-3y=1. 7y-20=1. 7y=21. y =3. x = 2(3)-4. x = 2. x*y =6.

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3 years ago

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Help!

AnnyKZ [126]

Answer:

2 stars and 3 points

Step-by-step explanation:

8+2=10

10+4=14

14+6=20

the value increases by 2 each time.

12+3=15

15+6=21

21+9=30

this value is based on multiples of three.

Good luck

7 0

3 years ago

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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; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 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