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2 years ago

5

B. y =

1 answer:

OleMash [197]2 years ago

5 0

Answer:

2 4 23

Step-by-step explanation:

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6. Using the discriminant, determine the value of k that will give 1 solution (i.e. discriminant equals zero) y = kx²-4x + 4

Gwar [14]

Answer:

k = 1

Step-by-step explanation:

<u>Discriminant</u>

$b^2-4ac\quad\textsf{when }\:ax^2+bx+c=0$

$\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}$

$\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}$

$\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}$

As we need to determine the value of k that will give <u>one solution</u>, set the <u>discriminant to zero</u>.

<u>Given equation</u>:

$y=kx^2-4x+4$

Therefore:

a = k
b = -4
c = 4

Substitute these values into the discriminant and solve for k:

$\begin{aligned}b^2-4ac & = 0\\\implies (-4)^2-4(k)(4) & = 0\\16-16k & = 0\\16k & = 16\\\implies k & = 1\end{aligned}$

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1 year ago

Lkn f.vwjn kowqnbvkjejwajvj dnrwjojni Helpppp.

chubhunter [2.5K]

The equation is 18+2.50h=12+3.75h

The total cost of the services will be equal at 4.8 Hours

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2 years ago

Anyone know the movie queen of the south well i wanna be Teresa shes so powerful

juin [17]

Answer:

39

Step-by-step explanation:

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2 years ago

Read 2 more answers

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

2 years ago

What is the factored form of x^2 -4x-12

denis-greek [22]

How about this?

(x - 6) x (x + 2)

8 0

2 years ago