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2 years ago

Please help me im really bad only higher grades can answer

Mathematics

2 answers:

NARA [144]2 years ago

6 0

A positive the numbers behind 0 are negative

myrzilka [38]2 years ago

3 0

Positive sorry if it’s wrong good luck

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fred can complete a work in 6 days and Paul can complete a work in 3 days. if fred and Paul work together, how many days can the

defon

Answer:

I think its 3 or 2 days bc if paul can get two works done by the time fred can get one done then the days should be divided.

Step-by-step explanation:

3 0

2 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that 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.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

2 years ago

the three median of a triangle intersect as a point. which measurements could represent the segments of one of the medians? 2 an

topjm [15]

2 & 3 is your answer

6 0

3 years ago

The equation of the parabola is (y-1)^2 = 16(x+3)

Mandarinka [93]

We know that

if the y terms is squared so it opens left or right

Is a Horizontal Parabola, the equation is of the form
(y-k)²=4(p)(x-h)
vertex is (h,k)
and
p is distance from focus to vertex, also distance from vertex to directrix
if p>0, then it opens to the right and directrix is to the left of vertex
if p<0, then it opens to the left and directrix is to the right of vertex

so
(y-1)²=4(4)(x-(-3))
vertex is (-3,1)
4>0 so directrix is to left of vertex
left is in x direction
-3-4=-7
directrix is x=-7

the answer is
x=-7

see the attached figure

5 0

2 years ago

Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.

Mrac [35]

Since molly's solution tally's with the given solution, hence <em>Molly's solution is correct.</em>

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

$\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}$

This partial product can also be solved as shown below:

$93 \times 51 = (90+3)\times (50+1)$

Applying the distributive law:

$93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743$

Since molly's solution tally's with the given solution, hence <em>Molly solution is correct.</em>

Learn more about partial product at: brainly.com/question/24716925

5 0

2 years ago