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3 years ago

Plz help me well mark brainliest!..

Mathematics

1 answer:

VladimirAG [237]3 years ago

5 0

9514 1404 393

Answer:

(a) (-3/4, 1/2)

Step-by-step explanation:

There are 4 grid lines between 0 and 1 in each direction, so each grid line represents 1/4 unit. Two of them is 1/2 unit; 3 of them is 3/4 unit. The x-dimension is listed first in the ordered pair.

Point S has coordinates (-3/4, 1/2).

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How do you do this? Because i don't know how to do it :>

Sindrei [870]

Answer:

£ 7.95

Step-by-step explanation:

Let the total money = £ x

Money spend on drinks and cake = 1.45 + 1.20 = £ 2.65

Money left with her = (2/3) of x

So, money spend = (1/3) of x

$\dfrac{1}{3}*x= 2.65\\\\\\x=2.65*3\\\\$

x = £ 7.95

7 0

2 years ago

A^3 b^2/a^-1 b^-3 please help!!!!!!!!!!!!!!!!!!!!!!

SashulF [63]

Answer:

Hope the picture will help you.............

8 0

2 years ago

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav

pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

If Z > 2, the result is considered significantly high.

If $np \geq 10$ and $n(1-p) \geq 10$ , the binomial distribution can be approximated to the normal with:

$\mu = np$

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

In this problem:

943 peas, thus, $n = 943$
3/4 probability of being red, thus $p = \frac{3}{4} = 0.75$ .

Applying the approximation:

$\mu = np = 943(0.75) = 707.25$

$\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297$

Item a:

Using continuity correction, this probability is $P(X \geq 717 - 0.5) = P(X \geq 716.5)$ , which is 1 subtracted by the p-value of Z when X = 716.5.

Then:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{716.5 - 707.25}{13.297}$

$Z = 0.7$

$Z = 0.7$ has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0

2 years ago

Three bags of sweets weigh 6 3/4 kg. Two of them have the same weight and the third bag is heavier than each of the bags of equa

Softa [21]

Answer: the weight of each bags are 1.85kg, 1.85kg and 3.05 kg

Step-by-step explanation:

Total weight of the three bags is

6 3/4 kg. Converting to decimal, it becomes 6.75 kg.

Two of them have the same weight. Let x represent the bags of equal weight. Their total weight would be 2x. The third bag is heavier than each of the bags of equal weight by 1 1/5 kg. Converting to decimal, it becomes 1.2kg. It means that the weight of the third bag would be

x + 1.2

Therefore, for the three bags,

2x + x + 1.2 = 6.75

3x = 6.75 - 1.2 = 5.55

x = 5.55/3 = 1.85

Weight of the third bag is

1.85 + 1.2 = 3.05kg

8 0

3 years ago

Nvm I do not need this

NemiM [27]

Wdym? Thx for the pts tho lol

6 0

2 years ago