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3 years ago

Which of the following represents this function written in intercept form?

2 answers:

7 0

Let us first solve the equation given in the question.

<span>y= x^2 - 8x + 12
= x^2 - 2x - 6x + 12
= x(x - 2) - 6(x - 2)
= (x - 2)(x - 6)

From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the second option or option "b". It actually </span><span>represents this function written in intercept form. I hope the procedure is clear enough for you to understand.</span>

mel-nik [20]3 years ago

6 0

Answer:

Like the other person said, the answer is b

Step-by-step explanation:

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A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude

I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

( a ) = 35 cm

( b ) = 54 cm

( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

<u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

$\sf{ \pmb { \longrightarrow \: \sqrt{s(s - a)(s - b)(s - c)} }}$

where ( s ) :

$\sf \longrightarrow s = \dfrac{a + b + c}{2}$

Therefore, for the given triangle first we will calculate ( s )

$\begin {aligned}\quad & \quad \longmapsto \sf s = \dfrac{a + b + c}{2} \\ & \quad \longmapsto \sf s = \dfrac{35 + 54 + 61}{2} \\ & \quad \longmapsto \sf s = \dfrac{150}{2} \\ & \quad \longmapsto \sf s = 75cm \end{aligned}$

Now, Area of triangle will be:

$\begin{aligned}&:\implies \sf\quad \sf \: A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \: A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)} \\&:\implies \sf\quad \sf \: A = \sqrt{75 \times 40 \times 21 \times 14} \\ &:\implies \sf\quad \sf \: A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5} \\ &:\implies \sf\quad \sf \: A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5} \\ &:\implies \sf\quad \sf \: A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf A =939.15 {cm}^{2} }} \end{aligned}$

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

$\begin{aligned}&:\implies \sf\quad \sf \: Area =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times b \times h =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times 61 \times h = 939.15 \\&:\implies \sf\quad \sf \: h =939.15 \times \dfrac{2}{61} \\&:\implies \sf\quad \sf \: h = \dfrac{1818.3}{61} \\ &:\implies \sf\quad \boxed{ \pmb{\sf \: h =30.79 \: (approx)}} \end{aligned}$

3 0

2 years ago

Read 2 more answers

A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select

Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; $\mu$ = 57,800 and $\sigma$ = 750.

Let X = randomly selected element of the population

The z probability is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{58000-57800}{750}$ ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{57000-57800}{750}$ ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

= 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = $\frac{\sigma}{\sqrt{n} }$ = $\frac{750}{\sqrt{100} }$ = 75

The z probability is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{58000-57800}{\frac{750}{\sqrt{100} } }$ ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{57000-57800}{\frac{750}{\sqrt{100} } }$ ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

7 0

3 years ago

A standard pair of six-sided dice is rolled. What is the probability of rolling a sum less than 7? Express your answer as a frac

zzz [600]

Answer:

0.4167 is the probability of rolling a sum less than 7.

Step-by-step explanation:

We are given the following in the question:

Event: A standard pair of six-sided dice is rolled

A: rolling a sum less than 7

Sample space:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

Formula:

$\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$P(A) = \dfrac{15}{36} = 0.4167$

0.4167 is the probability of rolling a sum less than 7.

8 0

3 years ago

F(1) = 4, f(n) = (−9) · f(n − 1) + 18

8090 [49]

f,n=4 and 27/20

this is the answer

3 0

3 years ago

Can someone help me with this. Will Mark brainliest.

Annette [7]

False hope that helps

3 0

3 years ago

Read 2 more answers