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3 years ago

Is a reflection of (-7,2) across the y-axis is at (7,-2)

Mathematics

1 answer:

Readme [11.4K]3 years ago

8 0

The reflection of (-7, 2) across the y-axis is (-7, -2) (hope this helps ;3)

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I will give lots of points please help

aalyn [17]

Answer:

a) 81π in³

b) 27 in³

c) divide the volume of the slice of cake by the volume of the whole cake

d) 10.6%

e) see explanation

Step-by-step explanation:

The cake can be modeled as a <u>cylinder </u>with:

diameter = 9 in
height = 4 in

$\sf Radius=\dfrac{1}{2}diameter \implies r=4.5\:in$

$\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}$

$\begin{aligned}\sf \implies \textsf{Volume of the cake} & =\pi (4.5)^2(4)\\ & = \sf \pi (20.25)(4)\\ & = \sf81 \pi \:\: in^3\end{aligned}$

$\begin{aligned}\textsf{Circumference of the cake} & = \sf \pi d\\& = \sf 9 \pi \:\:in\end{aligned}$

If each slice of cake has an arc length of 3 in, then the volume of each slice is 3/9π of the entire volume of the cake.

$\begin{aligned}\implies \textsf{Volume of slice of cake} & = \sf \dfrac{3}{9 \pi} \times \textsf{volume of cake}\\\\& = \sf \dfrac{3}{9 \pi} \times 81 \pi\\\\& = \sf \dfrac{243 \pi}{9 \pi}\\\\& = \sf 27\:\:in^3\end{aligned}$

The volume of each slice of cake is 27 in³.

The volume of the whole cake is 81π in³.

To calculate the probability that the first slice of cake will have the marble, divide the volume of a slice by the volume of the whole cake:

$\begin{aligned}\implies \sf Probability & = \sf \dfrac{27}{81 \pi}\\\\& = \sf 0.1061032954...\\\\ & = \sf 10.6\% \:\:(1\:d.p.)\end{aligned}$

Probability is approximately 10.6% (see above for calculation)

If the four slices of cake are cut and passed out <em>before </em>anyone eats or looks for the marble, the probability of getting the marble is the same for everyone. If one slice of cake is cut and checked for the marble before the next slice is cut, the probability will increase as the volume of the entire cake decreases, <u>until the marble is found</u>. So it depends upon how the cake is cut and distributed as to whether Hattie's strategy makes sense.

3 0

2 years ago

People attending a football game either supported the home team or the visiting team. if 1369 of the people attending supported

Masja [62]

Given the number of the people attending the football game, the percentage supporters for the home team is 37%.

<h3>What is Percentage?</h3>

Percentage is simply number or ratio expressed as a fraction of 100.

It is expressed as;

Percentage = ( Part / Whole ) × 100%

Given the data in the question;

Number of home team supporters nH = 1369

Number of visting team supporters nV = 2331

Percentage of supporters for home team PH = ?

For we determine the total number of people attending the football game;

nT = nH + nV

nT = 1369 + 2331

nT = 3700

Now, using the perecentage formula above, we find the Percentage of supporters for home team PH

Percentage = ( Part / Whole ) × 100%

PH = ( nH/ nT) × 100%

PH = ( 1369/ 3700) × 100%

PH = 0.37 × 100%

PH = 37%

Therefore, given the number of the people attending the football game, the percentage supporters for the home team is 37%.

Learn more about Percentages here: brainly.com/question/24159063

#SPJ1

6 0

2 years ago

Suppose that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage

Zigmanuir [339]

Answer:

We conclude that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%.

Let p = <u><em>population percentage of blue candies</em></u>

So, Null Hypothesis, $H_0$ : p = 29% {means that the percentage of blue candies is equal to 29%}

Alternate Hypothesis, $H_A$ : p $\neq$ 29% {means that the percentage of blue candies is different from 29%}

The test statistics that will be used here is <u>One-sample z-test for</u> <u>proportions</u>;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of blue coloured candies = 28%

n = sample of colored candies = 100

So, <u><em>the test statistics</em></u> = $\frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }$

= -0.22

The value of the z-test statistics is -0.22.

<u>Also, the P-value of the test statistics is given by;</u>

P-value = P(Z < -0.22) = 1 - P(Z $\leq$ 0.22)

= 1 - 0.5871 = 0.4129

Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29%.

4 0

3 years ago

SOMEONE PLEASE HELP<br> find the greatest common factor <br> 14g5h2, 56g3h4, 77g3h2

andrezito [222]

Answer:

$\large\boxed{GCF(14g^5h^2,\ 56g^3h^4,\ 77g^3h^2)=g^3h^2}$

Step-by-step explanation:

$14g^5h^2=2\cdot7\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot g\cdot g\cdot \boxed{h}\cdot \boxed{h}\\\\56g^3h^4=2\cdot2\cdot2\cdot7\cdot\boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{h}\cdot \boxed{h}\cdot h\cdot h\\\\77g^3h^2=7\cdot11\cdot\boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{h}\cdot \boxed{h}\\\\GCF(14g^5h^2,\ 56g^3h^4,\ 77g^3h^2)=\boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{h}\cdot \boxed{h}=g^3h^2$

8 0

3 years ago

George has $15. He would like to know if he has enough money to see a movie ($9.00) and buy a pretzel ($2.65), a drink ($1.35),

olga nikolaevna [1]

So George did't put parentheses around total snack amount to subtract his $15.He would need the total purchases to subtract amount by $15.

7 0

4 years ago

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