Suppose a stock is selling on a stock exchange for 6 3/4 dollars per share. If the price increases 3/4 per share, what is the new price of the stock? Leave your answer as a proper fraction or mixed number.

Answers

Answer 1

Answer:

The new price of the stock which increases 3/4 per share is $ 7 1/2.

Given data:

A stock is selling on a stock exchange for 6 3/4 dollars per share.

If the price of the stock increases by 3/4 dollars per share, you can add 3/4 to the current price of 6 3/4 dollars per share:

6 3/4 + 3/4

To add the whole numbers and fractions separately:

6 + 0 + 3/4 + 3/4

Adding the whole numbers and fractions:

6 + 1 1/2

A = 7 1/2

Hence, the price of the new stock is $ 7 1/2

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Answer 2

Answer:

Step-by-step explanation:

If the price increases 3/4 per share, the new price of the stock =

6¾ + ¾ =

27 /4 + ¾ =

30 /4 =

15/2or7½

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Round 369 to the nearest ten. Enter your answer in the box below.

Answers

Answer:

370

Step-by-step explanation:

nearest ten would be 370

Answer:

370

Step-by-step explanation:

just round the 9 into a 10

if your looking for the square root of this rounded its 19.20 (sorry if that's not what you needed I was confused)

The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.45, the analogous probability for the second signal is 0.5, and the probability that he must stop at at least one of the two signals is 0.9.Required:
a. What is the probability that he must stop at both signals?
b. What is the probability that he must stop at the first signal but not at the second one?
c. What is the probability that he must stop at exactly one signal?

Answers

Answer: a. 0.05

b. 0.40

c. 0.85

Step-by-step explanation:

Let F= Event that a certain motorist must stop at the first signal.

S = Event that a certain motorist must stop at the second signal.

As per given,

P(F) = 0.45 , P(S) = 0.5 and P(F or S) = 0.9

a. Using general probability formula:

P(F and S) =P(F) + P(S)- P(F or S)

= 0.45+0.5-0.9

= 0.05

∴ the probability that he must stop at both signals = 0.05

b. Required probability = P(F but (not s)) = P(F) - P(F and S)

= 0.45-0.05= 0.40

∴ the probability that he must stop at the first signal but not at the second one =0.40

c. Required probability = P(exactly one)= P(F or S) - P(F and S)

= 0.9-0.05

= 0.85

∴ the probability that he must stop at exactly one signal = 0.85

Final answer:

The probability of stopping at both signals is 0.225, the probability of stopping at the first one but not the second one is 0.225. The probability of stopping at exactly one signal is 0.675.

Explanation:

The probability theory can be used to answer these questions. The probabilities of stopping at various traffic signals can be calculated using some assumptions about the independence of the events.

The probability of the motorist having to stop at both signals can be found by multiplying the individual probabilities, assuming that these are independent events. So, P(stop at both signals) = P(stop at first signal) * P(stop at second signal) = 0.45 * 0.5 = 0.225.
The probability of him stopping at the first signal but not at the second one is again found by multiplying the probability of stopping at the first by the probability of not having to stop at the second. Therefore, P(stop at first, not at second) = 0.45 * (1 - 0.5) = 0.225.
To find the probability that the driver must stop at exactly one signal, we can subtract the probability of stopping at both signals from the probability of stopping at least one signal. So, P(stop at one signal) = P(stop at at least one signal) - P(stop at both signals) = 0.9 - 0.225 = 0.675.

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(Circle one) A negative decimal multiplied by a positive decimal then multiplied again by a negative decimal will give a positive/negative answer.

Answers

Positive

Step-by-step explanation:

A positive and a negative add up to a negative so if you add the sum of the first two then add the negative it will be positive

What is the equation of the line through (-6,-5) and (-4,-4) in slope-intercept form?

Answers

Answer:

y = 1/2x − 2

Pls answer asapA traveler has 9 pieces of luggage.

How many ways can the traveler select 3 pieces of luggage for a trip?

Responses

84

84

504

504

60,480

60,480

362,880

Answers

Answer:

Step-by-step explanation:

There are 9 pieces of luggage.

We can select the first piece 9 different ways, the second piece 8 different ways, since there are only 8 pieces left. The third piece can be selected 7 different ways.

9*8*7

504

But order doesn't matter since we are taking 3 pieces.

Divide by (3*2*1)

504/6 = 84

There are 84 different combinations of luggage

Answer:

Step-by-step explanation:

Trust me

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,000 will be accepted. Assume that the competitor's bid x is a is a random variable that is uniformly distributed between $10,000 and $15,000.a. Suppose you bid $12,000. What is the probability that your bid will be accepted? (please show calculations)
b. Suppose you bid $14,000. What is the probability that your bid will be accepted? (please show calculations)
c. What amount should you bid to maximize the probability that you get the property? (please show calculations)d. Suppose you know someone who is willing to pay you $16,000 for the property. Would you consider bidding less than the amount in part (c)? Why or why not?

Answers

Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

The interval of the accepted bidding is $[ {\rm{\$ 10,000 , \$ 15,000}]$ , where b = $15000 and a = $10000.

The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

$\begin{array}{c}\nP\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\n\n = 1 - \int\limits_(12000)^(15000) {\frac{1}{{15000 - 10000}}} dx\n\n = 1 - \int\limits_(12000)^(15000) {\frac{1}{{5000}}} dx\n\n = 1 - \frac{1}{{5000}}\left[ x \right]_(12000)^(15000)\n\end{array}$

$=1- ([15000-12000])/(5000)\n\n=1-0.6\n\n=0.4$

(b) The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

$\begin{array}{c}\nP\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\n\n = 1 - \int\limits_(14000)^(15000) {\frac{1}{{15000 - 10000}}} dx\n\n = 1 - \int\limits_(14000)^(15000) {\frac{1}{{5000}}} dx\n\n = 1 - \frac{1}{{5000}}\left[ x \right]_(14000)^(15000)\n\end{array} P(X<14,000)=1-P(X>14000)$

$=1- ([15000-14000])/(5000)\n\n=1-0.2\n\n=0.8$

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,

The interval of the accepted bidding is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The interval of the given bidding is [$10,000,$15,000].

$\begin{array}{c}\nf\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\n\n{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\n\n{\rm{ = 1}}\n\end{array}$

(d) The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

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