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lozanna [386]
4 years ago
5

Which should I choose​

Mathematics
1 answer:
Luden [163]4 years ago
3 0

Answer:

u = \dfrac{7-\frac{1}{2}at^2}{t}

Step-by-step explanation:

7 = \frac{1}{2}at^2+ut \\\\ ut = 7-\frac{1}{2}at^2 \\ \\ u = \dfrac{7-\frac{1}{2}at^2}{t}

You might be interested in
a normal distribution has a mean of 56 and a standard deviation of 8. Find the percentage of data values that are in the given i
vekshin1

Answer:

12) Between 40 and 64 = 0.815

13) Between 32 and 40 = 0.0235

14) Between 56 and 64 = 0.34

15) At most 56 = 0.515

16) At least 72 = 0.025

17) At most 64 = 0.855

Explanation:

To answer this, we will convert each of the values into their standardized form to make this easier.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

x = Each value

μ = Mean = 56

σ = Standard deviation = 8

12) Between 40 and 64

For 40,

z = (x - μ)/σ = (40 - 56)/8 = (-16/8) = -2

For 64

z = (x - μ)/σ = (64 - 56)/8 = (8/8) = 1

So,

Between 40 and 64 = Between -2 and 1

From the curve, noting that the central point is the mean, with standard score of 0, the lines before it move in step of 1 standard deviation towards the negative side, that is, -1, -2, etc. And the lines before the central point move towards the positive side, that is, 1, 2, 3, etc.

So,

Between -2 and 1 = 0.135 + 0.34 + 0.34 = 0.815

13) Between 32 and 40

For 32,

z = (x - μ)/σ = (32 - 56)/8 = (-24/8) = -3

For 40,

z = (x - μ)/σ = (40 - 56)/8 = (-16/8) = -2

So,

Between 32 and 40 = Between -3 and -2 = 0.0235

14) Between 56 and 64

For 56,

z = (x - μ)/σ = (56 - 56)/8 = (0/8) = 0

For 64,

z = (x - μ)/σ = (64 - 56)/8 = (8/8) = 1

Between 56 and 64 = Between 0 and 1 = 0.34

15) At most 56

For 56,

z = (x - μ)/σ = (56 - 56)/8 = (0/8) = 0

At most 56 = At most 0 = 0.0015 + 0.0235 + 0.15 + 0.34 = 0.515

All the regions before z = 0)

16) At least 72

For 72,

z = (x - μ)/σ = (72 - 56)/8 = (16/8) = 2

At least 72 = At least 2 = 0.0235 + 0.0015 = 0.025

(All the regions from z = 2 to the end)

17) At most 64

For 64,

z = (x - μ)/σ = (64 - 56)/8 = (8/8) = 1

At most 64 = At most 1 = 0.0015 + 0.0235 + 0.15 + 0.34 + 0.34 = 0.855

(All the regions before z = 1)

Hope this Helps!!!

4 0
3 years ago
If the ratio of a circles sector is 7/16 what is the measure of the sectors arc
Whitepunk [10]
Assuming that you mean that the sector is 7/16th of the area, we know that there are 360 degrees in a circle. To find 7/16th of that, we can simply multiply 360 by 7/16 to get 157.5 as the measure of the sector's arc in degrees
8 0
3 years ago
The standard equation of a circle with center (h.k) and radius r is (% - hy * (y - Ky = p. Which
zlopas [31]

Answer:

YOUR A DOG ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3 0
3 years ago
A certain bowler can bowl a strike 85 % of the time. What is the probability that she ​a) goes three consecutive frames without
Artist 52 [7]

Answer:

a) 0.34% probability that she goes three consecutive frames without a​ strike.

b) 1.91% probability that she her first strike in the third ​frame

c) 99.66% probability that she has at least one strike in the first three ​frames.

d) 14.22% probability that she bowls a perfect game.

Step-by-step explanation:

For each frame, there are only two possible outcomes. Either there is a strike, or there is not. The probability of a strike happening in a frame is independent of other frames. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A certain bowler can bowl a strike 85 % of the time.

This means that p = 0.85

a) goes three consecutive frames without a​ strike?

This is P(X = 0) when n = 3. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.85)^{0}.(0.15)^{3} = 0.0034

0.34% probability that she goes three consecutive frames without a​ strike.

​b) makes her first strike in the third ​frame?

No strike during the first two(with a 15% probability)

Strike during the third(85% probability). So

P = 0.15*0.15*0.85 = 0.0191

1.91% probability that she her first strike in the third ​frame

c) has at least one strike in the first three ​frames? ​

Either there are no strikes, or there is at least one strike. The sum of the probabilities of these events is 100%.

From a), 0.34% probability that she goes three consecutive frames without a​ strike.

100 - 0.34 = 99.66

99.66% probability that she has at least one strike in the first three ​frames.

d) bowls a perfect game​ (12 consecutive​ strikes)?

This is P(X = 12) when n = 12. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 12) = C_{12,12}.(0.85)^{12}.(0.15)^{0} = 0.1422

14.22% probability that she bowls a perfect game.

3 0
4 years ago
How many pair of whole number have a sum of 110
Zielflug [23.3K]
1,2,5,10,11,22,55,110
4 0
4 years ago
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