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2 years ago

The circle below is centered at the point (-3/4) and has a radius of length 3 what is its equation??

2 answers:

6 0

{x-(-3/4)}^2=3^2
(x+3/4)^2=9
{(4x+3)/4}^2=9
(4x+3)^2/16=9
(4x+3)^2=9*16
16x^2+2*4x*3+3^2=144
16x^2+24x+9=144
16x^2+24x+9-144=0
16x^2+24x-135=0

Juli2301 [7.4K]2 years ago

4 0

Answer:

$(x+3)^2+(y-4)^2=9$

Step-by-step explanation:

We are given a circle which is centered at the point (-3,4) and has a radius of length 3.

We have to find the equation of the circle.

The equation of circle centered at (a,b) and radius r is:

$(x-a)^2+(y-b)^2=r^2$

Here a=-3 , b=4 and r=3

Hence, equation of circle is:

$(x+3)^2+(y-4)^2=3^2$

$(x+3)^2+(y-4)^2=9$

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2 years ago

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At an intersection, the red-light times are normally distributed with a mean time of 3 minutes and a standard deviation of 0.25

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Percent of red lights last between 2.5 and 3.5 minutes is 95.44% .

Step-by-step explanation:

Step 1: Sketch the curve.

The probability that 2.5<X<3.5 is equal to the blue area under the curve.

Step 2:

Since μ=3 and σ=0.25 we have:

P ( 2.5 < X < 3.5 ) =P ( 2.5−3 < X−μ < 3.5−3 )

⇒ P ( (2.5−3)/0.25 < (X−μ)/σ < (3.5−3)/0.25)

Since, Z = (x−μ)/σ , (2.5−3)/0.25 = −2 and (3.5−3)/0.25 = 2 we have:

P ( 2.5<X<3.5 )=P ( −2<Z<2 )

Step 3: Use the standard normal table to conclude that:

P ( −2<Z<2 )=0.9544

Percent of red lights last between 2.5 and 3.5 minutes is $0.9544(100) = 95.44$ % .

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2 years ago

A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored

pickupchik [31]

Answer:

We conclude that there is an equal or larger proportion of Republicans in favor of lowering the standards.

Step-by-step explanation:

The complete question is: A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored lowering environmental standards so that high-sulfur coal could be burned in coal-fired power plants. The results were:

Number sampled: 1,000 (republican) , 800 (democrats)

Number in favor: 200 (republican) , 168 (democrats)

At the 0.02 level of significance, can we conclude that there is a larger proportion of Democrats in favor of lowering the standards? Determine the p-value.

Let $p_1$ = proportion of Republicans in favor of lowering the standards.

$p_2$ = proportion of Democrats in favor of lowering the standards.

SO, Null Hypothesis, $H_0$ : $p_1 \geq p_2$ {means that there is an equal or larger proportion of Republicans in favor of lowering the standards}

Alternate Hypothesis, $H_A$ : $p_1 < p_2$ {means that there is a larger proportion of Democrats in favor of lowering the standards}

The test statistics that would be used here Two-sample z-test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of Republicans in favor of lowering the standards = $\frac{200}{1000}$ = 0.20

$\hat p_2$ = sample proportion of Democrats in favor of lowering the standards = $\frac{168}{800}$ = 0.21

$n_1$ = sample of Republicans = 1000

$n_2$ = sample of Democrats = 800

So, the test statistics = $\frac{(0.20-0.21)-(0)}{\sqrt{\frac{0.20(1-0.20)}{1000}+\frac{0.21(1-0.21)}{800} } }$

= -0.52

The value of z test statistics is -0.52.

Now, P-value of the test statistics is given by the following formula;

P-value = P(Z < -0.52) = 1 - P(Z $\leq$ 0.52)

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Now, at 0.02 significance level, the z table gives a critical value of -2.054 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.52 > -2.054, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is an equal or larger proportion of Republicans in favor of lowering the standards.

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2 years ago

A root beer factory produces 132554 cases in 100 days. About how many cases does the factory produce in 1 day?

Juli2301 [7.4K]

It is 56 cases hoped i help (:

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2 years ago

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