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3 years ago

Find the area with Heron's formula pls help !

Mathematics

2 answers:

KatRina [158]3 years ago

8 0

Answer:

3320in²

Step-by-step explanation:

heron formula = √s(s-a)(s-b)(s-c)

when s = (a+b+c)/2

s = (16+35+25)/2 = 76/2 = 38

heron formula = √s(s-a)(s-b)(s-c) = √78(78-16)(78-35)(78-25) = √78(62)(43)(53) = √11021244 = 3319.82589905≈3319.8in²(nearest tenth)

Mandarinka [93]3 years ago

5 0

Answer: Area = 180.6 m²

Step-by-step explanation:

Heron's formula for determining the area of a rectangle is expressed as

Area = √S(S - a)(S - b)(S - c)

Where

a,b and c represents the length of each side of the triangle

S represents perimeter/2

Perimeter = 16 + 35 + 25 = 76

S = 76/2 = 38

Applying Heron's formula, it becomes

Area = √38(38 - 16)(38 - 25)(38 - 35)

= √38 × 22 × 13 × 3

= √32604

= 180.6 m² rounded to the nearest tenth.

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3 years ago

What is the equation of the line in slope-intercept form with slope -3/2 and y-intercept -5?

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3 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

Francis is at the state fair. A fair concession stand is selling funnel cakes for $3.50 and deep-fried Oreos for $2.00. Write an

VladimirAG [237]

The equation that models the number of funnel cakes and Oreos he can buy is 3.50x + 2.0y = 42

Data given;

Cost of cakes = $3.50

Cost of Oreos = $2.00
The total amount spent = $42.00

<h3>What is the Equation</h3>

To solve this problem, we just need to write out an equation to show how he can spend $42.00 in the fair on Oreos and Cakes.

Let x represent the cakes

Let y represent the Oreos

The equation is thus;

$3.50x + 2.0y = 42.00$

The equation that shows the number of Cakes and Oreos can by is

3.50x + 2.0y = 42

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2 years ago

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Answer:

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