Y
X

Pls help me you’ll get 29 points

Answer:

15

Step-by-step explanation:

By order of operations, multiplication and division are done first, then the addition and subtraction. Remember, multiplication and division have the same precedence, as does addition and subtraction.

21*6 = 126

126/7 = 18

18 + 12 = 30

30 - 15 = 15

Answer:

15

Step-by-step explanation:

21 × 6 ÷ 7 + 12 - 15​

= 126 ÷ 7 + 12 - 15

= 18 + 12 - 15

= 30 - 15

= 15

Answer:

17 : 27

Step-by-step explanation:

x=3

y=5

4(3)+5 : 6(5)-3

= 12+5 : 30-3

= 17 : 27

Answer:

B

Step-by-step explanation:

they have the same intercepts

Answer:

AC = 377.19

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp /adj

tan 5 = 33/AC

AC tan 5 = 33

AC = 33/ tan 5

AC = 377.19

Answer:

Step-by-step explanation:

Answer:

3/8 x 5/8= 15/64

Step-by-step explanation:

Answer:

x=8

Step-by-step explanation:

Answer:

[tex]|F'H'| = 2 * |FH|[/tex]

Step-by-step explanation:

Given

[tex]E = (0,1)[/tex]             [tex]E' = (-1,2)[/tex]

[tex]F = (1,1)[/tex]             [tex]F' = (1,2)[/tex]

[tex]G = (2,0)[/tex]             [tex]G' =(3,0)[/tex]

[tex]H = (0,0)[/tex]            [tex]H' = (-1,0)[/tex]

[tex](x,y) = (1,0)[/tex] -- center

[tex]k = 2[/tex] --- scale factor

See comment for proper format of question

Required

Compare FH to F'H'

From the question, we understand that the scale of dilation from EFGH to E'F'G'H is 2;

Irrespective of the center of dilation, the distance between corresponding segment will maintain the scale of dilation.

i.e.

[tex]|F'H'| = k * |FH|[/tex]

[tex]|F'H'| = 2 * |FH|[/tex]

To prove this;

Calculate distance of segments FH and F'H' using:

[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

Given that:

[tex]F = (1,1)[/tex]             [tex]F' = (1,2)[/tex]

[tex]H = (0,0)[/tex]            [tex]H' = (-1,0)[/tex]

We have:

[tex]FH = \sqrt{(1- 0)^2 + (1- 0)^2}[/tex]

[tex]FH = \sqrt{(1)^2 + (1)^2}[/tex]

[tex]FH = \sqrt{1 + 1}[/tex]

[tex]FH = \sqrt{2}[/tex]

Similarly;

[tex]F'H' = \sqrt{(1 --1)^2 + (2 -0)^2}[/tex]

[tex]F'H' = \sqrt{(2)^2 + (2)^2}[/tex]

Distribute

[tex]F'H' = \sqrt{(2)^2(1 +1)}[/tex]

[tex]F'H' = \sqrt{(2)^2*2}[/tex]

Split

[tex]F'H' = \sqrt{(2)^2} *\sqrt{2}[/tex]

[tex]F'H' = 2 *\sqrt{2}[/tex]

[tex]F'H' = 2\sqrt{2}[/tex]

Recall that:

[tex]|F'H'| = 2 * |FH|[/tex]

So, we have:

[tex]2\sqrt 2 = 2 * \sqrt 2[/tex]

[tex]2\sqrt 2 = 2\sqrt 2[/tex] --- true

Hence, the dilation relationship between FH and F'H' is::

[tex]|F'H'| = 2 * |FH|[/tex]

Answer:NOTT !!  A segment in the image has the same length as its corresponding segment in the pre-image.

Step-by-step explanation:

Answer:

0.0287 = 2.87% probability the total of the 100 claims exceeds 1,530,000.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n instances of a normal variable:

For n instances of a normal variable, the mean is [tex]n\mu[/tex] and the standard deviation is [tex]s = \sigma\sqrt{n}[/tex]

Sum of normal variables:

When two normal variables are added, the mean is the sum of the means, while the standard deviation is the square root of the sum of the variances.

Group A follow a normal distribution with mean 10,000 and standard deviation 1,000. 50 claims of group A.

This means that:

[tex]\mu_A = 10000*50 = 500000[/tex]

[tex]s_A = 1000\sqrt{50} = 7071[/tex]

Group B follow a normal distribution with mean 20,000 and standard deviation 2,000. 50 claims of group B.

This means that:

[tex]\mu_B = 20000*50 = 1000000[/tex]

[tex]s_B = 2000\sqrt{50} = 14142[/tex]

Distribution of the total of the 100 claims:

[tex]\mu = \mu_A + \mu_B = 500000 + 1000000 = 1500000[/tex]

[tex]s = \sqrt{s_A^2+s_B^2} = \sqrt{7071^2+14142^2} = 15811[/tex]

Find the probability the total of the 100 claims exceeds 1,530,000.

This is 1 subtracted by the p-value of Z when X = 1530000. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{1530000 - 1500000}{15811}[/tex]

[tex]Z = 1.9[/tex]

[tex]Z = 1.9[/tex] has a p-value of 0.9713

1 - 0.9713 = 0.0287

0.0287 = 2.87% probability the total of the 100 claims exceeds 1,530,000.

Answer:

His least score for him in the fourth paper has to be 76.

Step-by-step explanation:

Given that Michael has an average of 68% in his 3 papers but that is below the pass mark of 70%, to determine what must be his least score in the fouth paper to enable him pass the following calculation must be performed:

(70 x 4) - (68 x 3) = X

280 - 204 = X

76 = X

Therefore, his least score for him in the fourth paper has to be 76.

Answer:

15. 52

16. 6

17. 59

18. 11

Step-by-step explanation:

Answer:

Muhammad lives 8 km away from the school.

Hita lives 4 km away from the school.

Step-by-step explanation:

First of all, find a number that, when you double that number and add both numbers, you will get 12. That number is 4. So double 4 and get 8. Then add both to get 12.

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\large\textsf{6!}\\\large\textsf{= 6}\times\large\textsf{5}\times\large\textsf{4}\times\large\textsf{3}\times\large\textsf{2}\times\large\textsf{1}\\\large\textsf{6(5) = \bf 30}\\\large\textsf{= 30}\times\large\textsf{4}\times\large\textsf{3}\times\large\textsf{2}\times\large\textsf{1}\\\large\textsf{30(4) = \bf 120}\\\large\textsf{= 120}\times\large\textsf{3}\times\large\textsf{2}\times\textsf{1}\\\large\textsf{120(3) = \bf 360}\\\large\textsf{= 360}\times\large\textsf{2}\times\large\textsf{1}[/tex]

[tex]\large\textsf{360(2) = \bf 720}\\\large\textsf{720}\times\large\textsf{1}\\\large\textsf{= \bf 720}[/tex]

[tex]\boxed{\boxed{\huge\textsf{Therefore, your answer is: \bf 720}\huge\textsf{ (option B)}}}\huge\checkmark[/tex]

[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

-2

-1

-2

Step-by-step explanation:

really ? this is a problem ? why ?

f(0) means the functional value for x = 0.

is x = 2 ? no.

so, automatically the other case applies, and f(0) = -2

f(2) means x=2

is x = 2 ? yes.

so that case applies, and f(2) = -1

f(5) means x=5

is x = 2 ? no.

so again, the case for x <> 2 applies, f(5) = -2

This question is incomplete, the complete question is;

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule.

What proportion of the students scored at least 23 points on this test, rounded to five decimal places?

Answer:

proportion of the students that scored at least 23 points on this test is 0.30850

Step-by-step explanation:

Given the data in the question;

mean μ = 22

standard deviation σ = 2

since test closely followed a Normal Distribution

let

Z = x-μ / σ      { standard normal random variable ]

Now, proportion of the students that scored at least 23 points on this test.

P( x ≥ 23 ) = P( (x-μ / σ) ≥  ( 23-22 / 2 )

= P( Z ≥ 1/2 )

= P( Z ≥ 0.5 )

= 1 - P( Z < 0.5 )

Now, from z table

{ we have P( Z < 0.5 ) = 0.6915 }

= 1 - P( Z < 0.5 ) = 1 - 0.6915 = 0.30850

P( x ≥ 23 ) = 0.30850

Therefore, proportion of the students that scored at least 23 points on this test is 0.30850

Answer:

[tex]x=-\frac{15}{382}[/tex]

Step-by-step explanation:

32 × 12x + 1 = 2x - 14

384x + 1 = 2x - 14

384x + 1 - 1 = 2x - 14 - 1

384x = 2x - 15

384x - 2x = 2x - 2x - 15

382x = - 15

382x ÷ 382 = - 15 ÷ 382

[tex]x=-\frac{15}{382}[/tex]

Answer:

Sample space = 36

P(sum of 6) = 5/36

Step-by-step explanation:

Number of faces on a dice = 6

The sample space, for a toss of 2 dice ; (Number of faces)^number of dice

Sample space = 6^2 = 6*6 = 36

Sum of numbers appearing on the dice = 6

The sum of 6 from the roll of two dice has 5 different outcomes ; Hence, required outcome = 5

Total possible outcomes = sample space = 36

Probability, P = required outcome / Total possible outcomes

P = 5 / 36

Probabilities are used to determine the chances of events

The given parameters are:

[tex]n=6[/tex] --- the faces of a six-sided die

[tex]r = 2[/tex] -- the number of dice

(a) The number of sample points

This is calculated as:

[tex]Sample = n^r[/tex]

So, we have:

[tex]Sample = 6^2[/tex]

Evaluate the exponent

[tex]Sample = 36[/tex]

Hence, the number of sample points is 36

(b) The probability that the sum of 6

See attachment for the sample space of the sum of two dice.

From the sample space, there are 5 outcomes where the sum is 6.

So, the probability is:

[tex]Pr = \frac{5}{36}[/tex] --- where 36 represents the number of sample points

Divide 5 by 36

[tex]Pr = 0.1389[/tex]

Hence, the probability that the sum of the numbers appearing on the dice is equal to 6 is 0.1389

Read more about probabilities at:

https://brainly.com/question/10707698

Answer:

4

Step-by-step explanation:

We want the cubed root of 64  

(64)^(1/3)  

(4*4*4) ^ (1/3)  

4  

Unless this is 3 * sqrt(64)  

then it would be  

3 sqrt(8*8)  

3 (8)  

24

Answer:

4s-9t-7

Step-by-step explanation:

multiply the negative one with all terms inside the bracket, since they are all unlike terms the answer remains the same

9514 1404 393

Answer:

  (-3, 3)

Step-by-step explanation:

The blanks are trying to lead you through the process of finding the point of interest.

__

The horizontal distance from T to S is 9 . (or -9, if you prefer)

The ratio you're trying to divide the line into is the ratio that goes in this blank:

Multiply the horizontal distance by 2/3 . (9×2/3 = 6)

Move 6 units left from point T.

The vertical distance from T to S is 6 .

Multiply the vertical distance by 2/3 . (6×2/3 = 4)

Move 4 units up from point T.

__

Point T is (3, -1) so 6 left and 4 up is (3, -1) +(-6, 4) = (3-6, -1+4) = (-3, 3). The point that is 2/3 of the way from T to S is (-3, 3).

Answer:

The answer is below

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, reflection, translation and dilation.

If a point A(x, y) is translated a units right and b units up, the new point is at A'(x + a, y + b).

If a point A(x, y) is reflected across the line y = -x, the new point is at A'(-y, -x).

If a point A(x, y) is rotated counterclockwise by 270 degrees, the new point is at A'(y, -x).

Let us assume that triangle ABC has vertices at A(-6, -1), B(-3, -3) and C(-1, -2).

If it is moved 5 units to the right and 2 units up, the new point is at A'(-1, 1), B'(1, -1) and C'(3, 0). If it is reflected across the line y = -x, the vertices are at A"(-1, 1), B"(1, -1) and C"(0, -3). If it is then rotated counterclockwise about the origin by 270°, the new point is at A'"(-1, -1), B"'(1, 1), C"'(3, 0)

Answer:

0.5665 = 56.65% probability of less than four twos.

Step-by-step explanation:

For each roll, there are only two possible outcomes. Either it is a two, or it is not a two. The probability of a roll ending up in a two is independent of any other roll, which means that the binomial probability distribution is used.

Binomial probability distribution

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

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

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A die is rolled 20 times

This means that [tex]n = 20[/tex]

One out of six sides is 2:

This means that [tex]p = \frac{1}{6} = 0.1667[/tex]

Probability of less than four twos:

This is:

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

So

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

[tex]P(X = 0) = C_{20,0}.(0.1667)^{0}.(0.8333)^{20} = 0.0261[/tex]

[tex]P(X = 1) = C_{20,1}.(0.1667)^{1}.(0.8333)^{19} = 0.1043[/tex]

[tex]P(X = 2) = C_{20,2}.(0.1667)^{2}.(0.8333)^{18} = 0.1982[/tex]

[tex]P(X = 3) = C_{20,3}.(0.1667)^{3}.(0.8333)^{17} = 0.2379[/tex]

So

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0261 + 0.1043 + 0.1982 + 0.2379 = 0.5665[/tex]

0.5665 = 56.65% probability of less than four twos.

Answer:

[tex]Given:[/tex] Δ ABC ≈ ΔDEF

[tex]therefor:[/tex] A(ΔABC)/A(ΔDEF)=(BC)²/(EF)²

⇒ 34/A(ΔDEF)=9²/(13.5)²

⇒34/A(ΔDEF)=81/182.25

⇒A(ΔDEF)=34×182.25/81

⇒Area of ΔDEF=76.5 cm²

----------------------------------

Hope it helps...

Have a great day!!!

Answer:

The correct answer is "76.98%".

Step-by-step explanation:

According to the question,

⇒ [tex]P(34000<x<46000) = P[\frac{34000-40000}{5000} <\frac{x- \mu}{\sigma} <\frac{46000-40000}{5000} ][/tex]

                                       [tex]=P(-1.2<z<1.2)[/tex]

                                       [tex]=P(z<1.2)-P(z<-1.2)[/tex]

                                       [tex]=0.8849-0.1151[/tex]

                                       [tex]=0.7698[/tex]

or,

                                       [tex]=76.98[/tex]%

AS IN THE PICTURE...........

Answer:

x = -3  or x = 3  or x = 6

Step-by-step explanation:

x3 – 6x2 – 9x + 54 = 0

There is no common factor to factor out. There are 4 terms. We try factoring by grouping. Factor a common factor out of the first two terms. Factor a common factor out of the last two terms.

x^2(x - 6) - 9(x - 6) = 0

x - 6 is a common factor, so we factor it out.

(x^2 - 9)(x - 6) = 0

x^2 - 9 is the difference of 2 squares, so we factor it.

(x + 3)(x - 3)(x - 6) = 0

x + 3 = 0  or  x - 3 = 0  or x - 6 = 0

x = -3  or x = 3  or x = 6

Answer:

x=6     x=3   x=-3

Step-by-step explanation:

x^3 – 6x^2 – 9x + 54 = 0

Factor by grouping

x^3 – 6x^2     – 9x + 54 = 0

x^2(x-6)    -9(x-6 ) =0

Factor out x-6

(x-6)(x^2 -9) =0

Notice x^2 -9 is the difference of squares

(x-6)(x-3)(x+3) = 0

Using the zero product property

x-6 =0    x-3 =0   x+3 =0

x=6     x=3   x=-3

...

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