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PSYCHO15rus [73]
2 years ago
14

Horn lengths of Texas longhorn cattle are normally distributed. The mean horn spread is 60 inches with a standard deviation of 4

.5 inches.
Calculate the range of horn lengths for the middle 68% of Texas longhorn cattle.
Mathematics
1 answer:
german2 years ago
8 0

Answer:

range is  between 55.5 to 64.5

Step-by-step explanation:

Horn lengths of Texas longhorn cattle are normally distributed. The mean horn spread is 60 inches with a standard deviation of 4.5 inches

68% is 1 standard deviation from mean

To get the range of 1 standard deviation we add and subtract standard deviation from mean

mean = 60

standard deviation = 4.5

60 - 4.5= 55.5

60+4.5 = 64.5

1 standard deviation is between 55.5 to 64.5

That is 68% range is  between 55.5 to 64.5

You might be interested in
Consider the following problem: A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four
Alina [70]

Answer:

Step-by-step explanation:

(a)

Suppose we came up with an ideology whereby we pick a value for the length including the length dividing the inside into 4 parts(5 parallel sides), then we can get the value for breath by using the following process.

Let assume the length of the rectangle is 50;

Then, the breath can be calculated as follows:

= 50 × 5 = 250   ( since the breath is divided into 5 parallel sides)

The fencing is said to be 950 ft

So, 950 - 250 = 700

Then divided by 2, we get:

= 700/2

= 350

So for the first diagram; the length = 50 and the breath = 350

The area = 50 × 350 = 17500 ft²

Now, let's go up a little bit.

If the length increase to 100;

Then 100 × 5 = 500

⇒ 950 - 500 = 450

⇒ 450/2 = 225

The area = 225 × 100 = 22500 ft²

Suppose the length increases to 150

Then 150 × 5 = 750

⇒ 950 - 750 = 200

⇒ 200/2 = 100

The area = 150 × 100 = 15000 ft²

The diagrams for each of the outline above can be seen in the image attached below.

(b) The diagram illustrating the general solution can be seen in the second image provided below.

(c) The expression for  the total area A in terms of both x and y is:

Area A = x×y

(d) Recall that:

The fencing is said to be 950 ft.

And the length is divided inside into 5 parallel sides;

Then:

5x + 2y = 950  (from the illustration in the second image below)

2y  = 950 - 5x

y = \dfrac{950}{2} - \dfrac{5}{2}x

y = 475- \dfrac{5}{2}x

(e)

From (c); replace the value of y in (d) into (c)

Then:

Area A = x×y

f(x)= x\times ( 475 -\dfrac{5}{2}x)

Open brackets

f(x)= ( 475 x-\dfrac{5}{2}x^2)

(f)

By differentiating what we have in (e)

f(x)= ( 475 x-\dfrac{5}{2}x^2)

f'(x)= ( 475 (1)-\dfrac{5}{2}(2x))

f'(x)= 475 -5x

\implies  475 = 5x

x = 475/5

x = 95

From (d):

y = 475- \dfrac{5}{2}x

y = 475- \dfrac{5}{2}(95)

y =237.5

∴

Area A = x × y

Area A = 95 × 237.5

Area A = 22562.5 ft²

5 0
2 years ago
The number of newly reported crime cases in a county in New York State is shown in the accompanying table, where x represents th
Novay_Z [31]

Answer:

Step-by-step explanation:

I use  84+ CE

stat edit, then fill in the #s

then

vars 5

then

2'nd stat plot, on

then, click stat

Click arrow 1 time to the left to get to Calc

then click (4)(LinReg(ax+b))

then click enter 5 times

(y=-25.31428571x+1000.285714

y=-25.3x+1000.3

now, lets use computer:

y=-25.31(543)+1000.3

y=-12743.03

round to the biggest whole number )

this doesn't really work, so I will put 1999, 2000, 2001, 2002, 2003, 2004 instead of 0, 1, 2, 3, 4, 5 and do the same thing

now I get

y=-25.31428571x+51603.54286

y=-25.3x+51603.5

now, lets use computer:

y=-25.3(543)+51603.5

y=37865.6

round to the biggest whole number:

y=37866

so, year 37866

7 0
2 years ago
A rectangular portrait measures 50cm by 70cm. It is surrounded by a rectangular frame of uniform width. If the area of the frame
snow_tiger [21]

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore,  length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be =  (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame -  area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

 3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula =\frac{-b\pm \sqrt{b^2-4ac}}{2a}, we get

=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}

x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)

x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)

We cant take negative value.

So, x=5\left(\sqrt{71}-6\right)=12.13

We could take approximately 12 cm.



7 0
2 years ago
"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul
Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0
2 years ago
Bryan earns $5.00 per lawn mowed and $5.00 per car washed. Let m be the number of lawns mowed and let c be the number of cars wa
Harrizon [31]

Answer:

5m+c

Step-by-step explanation:

4 0
2 years ago
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