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1 year ago

NEED HELP ASAP!!!!!!!!!!

Mathematics

1 answer:

OLga [1]1 year ago

8 0

Answer:

Question 1)

The graph of $y=\dfrac{2}{x}$ is attached to the answer.

The graph of such a functio lies in first and third quandrant such that the graph tends to 0 when x→∞ or when x→-∞ and graph tends to ∞ when x→0.

We can also show the values of this equation on a table:

 x-values  y-values

1/2 4

1 2

2 1

3 2/3

Question 2)

The graph of the equation $y=x^{2}+2x-3$ is attached to the answer.

we will get a upward parabola with this equation whose vetex is: (-1,-4).

The ordered pair on the graph could be shown with the help of a table:

x-values  y-values

-1 -4

0 -3

-3 0

1 0

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Erica can run 1 6 6 1 start fraction, 1, divided by, 6, end fraction of a kilometer in a minute. Her school is 3 4 4 3 start

Brut [27]

Her school is 2/3 miles away

2/3=4/6miles

So we need to find out how long it will take for her to run home from school...

School=4/6 miles

In 1 minutes she can run 1/6 miles

1min=1/6miles

In 2 minutes she can run 1/6+1/6 miles (1/6+1/6=2/6)

2min=2/6miles

3min=3/6miles

4min=4/6miles

It will take Erica 4 minutes to run 4/6 miles, so it'll take her 4 minutes to get home.

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2 years ago

In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t

victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that he or she would choose the pair of socks in the center position is $p =\frac{1}{5}$

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5

The mean is $\mu = 20$

The standard deviation is $\sigma=4$

The probability, $P =0.0002$

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability $Pe = 0.0003$

The probabilities $P \approx Pe$

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

$p =\frac{1}{5}$

$=0.2$

The mean of this distribution is mathematical represented as

$\mu = np$

substituting the value

$\mu = 100 * 0.2$

$\mu = 20$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np (1-p)}$

substituting the value

$= \sqrt{100 * 0,2 (1-0.2)}$

$\sigma=4$

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

$P=P(X \ge 34 )$

By standardizing the normal approximation we have that

$P(X \ge 34) \approx P(Z \ge z)$

Now z is mathematically evaluated as

$z = \frac{x-\mu}{\sigma }$

Substituting values

$z = \frac{34-20}{4}$

$=3.5$

So using the z table the $P(Z \ge 3.5)$ is 0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

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2 years ago

Flannery used 30 lilies and 78 roses to create six identical flower arrangements. Write an equation to describe the relationship

Volgvan

Number of lilies used= 30 lilies

Number of roses used= 78 roses

To make the six identical arrangements:

We need to divide lilies into six equal parts, as well as number of roses in six equal parts.

I=Number of lilies in each arrangement = $\frac{30}{6}=5$

r=Number of roses in each arrangement = $\frac{78}{6}=13$

Total number of flowers in arrangement (I + r)= 5 Lilies + 13 Roses=18 (Flowers)

The relationship between l, the number of lilies, and r, the number of roses:

→ 5 I = 13 r

$I=\frac{13 r}{5} , r =\frac{5I}{13}$

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1 year ago

a lunch stand makes a $.75 profit on each chef's salad and $1.20 profit on each caesar salad. On a typical weekday, it sells bet

tangare [24]

Answer:

50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.

Step-by-step explanation:

Suppose, the number of Chef's salad is $x$ and the number of Caesar salad is $y$

On a typical weekday, it sells between 40 and 60 Chefs salads and between 35 and 50 Caesar salads.

So, the two constraints are: $40\leq x\leq 60$ and $35\leq y\leq 50$

The total number sold has never exceed 100 salads. So, another constraint will be: $x+y\leq 100$

According to the graph of the constraints, the vertices of the common shaded region are: $(40,35), (60,35), (60,40), (50,50)$ and $(40,50)$ (Refer to the attached image for the graph)