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2 years ago

If f(t)= 3t^2-2, find f(k-2)

Mathematics

1 answer:

DedPeter [7]2 years ago

7 0

By plugging (k-2) in t, we can write that
$f(k-2)=3(k-2)^{2}-2=3(k^{2}-4k+4)-2=3 k^{2}-12k+12-2=$ $3 k^{2}-12k+10$

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A person riding a circular Ferris wheel sits 65 feet from the center. If they travel a distance of 92 feet during an initial rot

Temka [501]

We know that
the radius is 65 ft
circumference of a circle=2*pi*r
for r=65 ft
circumference of a circle=2*pi*65-----> 130*pi ft

if 2*pi (full circle) has a length of----------------> 130*pi ft
x------------------------------------------> 92 ft
x=92*2*pi/(130*pi)----------> x=1.4154 radians-------> x=1.42 radians

the answer is
1.42 radians

8 0

2 years ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as

lubasha [3.4K]

Answer:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

$V = l * b * h$

Make h the subject of the formula

$h = \frac{V}{lb}$

The surface area (S) of the aquarium is:

$S = lb + 2(lh + bh)$

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

$C = 7 * lb + 1 * 2(lh + bh)$

$C = 7lb + 2(lh + bh)$

$C = 7lb + 2h(l + b)$

Substitute $\frac{V}{lb}$ for h in the above equation

$C = 7lb + 2*\frac{V}{lb}(l + b)$

$C = 7lb + \frac{2V}{lb}(l + b)$

$C = 7lb + \frac{2V}{b} + \frac{2V}{l}$

Differentiate with respect to l and with respect to b

$C_l=7b - \frac{2V}{l^2}$ $=0$

$C_b=7l - \frac{2V}{b^2}$ $=0$

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

$7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}$

$7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}$

By comparison:

$l =b$

$C_l=7b - \frac{2V}{l^2}$ $=0$ becomes

$7l - \frac{2V}{l^2}=0$

$7l = \frac{2V}{l^2}$

Cross Multiply

$7l^3 = 2V$

Solve for l

$l^3 = \frac{2V}{7}$

$l = \sqrt[3]{\frac{2V}{7}}$

Recall that: $l =b$

$b = \sqrt[3]{\frac{2V}{7}}$

Also recall that:

$h = \frac{V}{lb}$

$h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}$

$h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}$

Apply law of indices

$h = \sqrt[3]{\frac{49V^3}{4V^2}}$

$h = \sqrt[3]{\frac{49V}{4}}$

The dimension that minimizes the cost of material of the aquarium is:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

8 0

2 years ago

140 fluid ounces increased by 45%

Leokris [45]

First divided 140 by 100 to find out how much 1 percent is
140 ÷ 100 = 1.4

multiply 1.4 by 45 to find 45 %; 1.4 x 45 = 63

then add 63, which is 45%, to 140, which is 100%, and you get 203, which is 140 + 45%, or 145%

please mark brainliest if this helped you. I hope it did!

4 0

2 years ago

In a box containing 25 cherries, 2 of them are rotten. Susan randomly picks cherries in the box. How many cherries should be pic

bezimeni [28]

Answer:

Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20

Step-by-step explanation:

assuming that each cherry is equally probable to be chosen , since each cherry is independent from the others and sampling is done without replacement , the random variable X= number of cherries that are rotten from the picked ones follows a hyper geometrical distribution , where

P(X=k)= C(M,k) * C(N-M, n-k) / C(N,n)

where

N= population size = 25

n= number of picks

M = total number of rotten cherries =2

k = number of rotten cherries picked =2

C( ) = combination

then

1/20=C(2,2)*C(25-2,n-2)/C(25,n) = 1 * (23!/(n-2)!*(25-n)! / (25!/(n!*(25-n)!

1/20 = n!/(n-2)! * 1/(24*25)

24*25/20 = n*(n-1)

n²-n-30 =0

n= (1 +√(1+4*1*30))/2 = 12/2= 6

n=6

then Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20

7 0

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)

= 1 - 0.6985 = 0.3015

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.

7 0

2 years ago