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-BARSIC- [3]
2 years ago
11

Roman completed the square to solve the equation 0=−x2+6x−15 0 = − x 2 + 6 x − 15 as follows. Step 1: x2−6x+15=0 x 2 − 6 x + 15

= 0 Step 2: x2−6x+15−15=0−15 x 2 − 6 x + 15 − 15 = 0 − 15 Step 3: x2−6x+9=−15+9 x 2 − 6 x + 9 = − 15 + 9 Step 4: (x−3)2=−6 ( x − 3 ) 2 = − 6 Step 5: x−3=±−6−−−√ x − 3 = ± − 6 Step 6: x−3−3=−3±−6−−−√ x − 3 − 3 = − 3 ± − 6 Step 7: x=−3±i6√ x = − 3 ± i 6 Did Roman make a mistake? If so, where?
Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
5 0

Answer:

yes

Step-by-step explanation:

at step 6, his meant to add 3 to both side not subtract

and at step 7, it's meant to be i√6 not i6

hope this helps:)

You might be interested in
9,6,3,0,-3,-6 work out the nth term (it decrease by the same amount each time)
Shkiper50 [21]

Answer:

-3n+11 or 3n+11 or 3n-11 one of those

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Shaan walks 2.5 meters per second. His brother Dhvan walks 1 meter
MissTica

Answer:

The race must be up to 29 meters for Dhvan to win.

Step-by-step explanation:

Since Shaan walks 2.5 meters per second, while his brother Dhvan walks 1 meter per second, and Dhvan wants to have a race, and Shaan knows that he walks faster, but he wants to give his brother a head start of 45 meters, so it doesn't seem that he is allowing him to win, to determine how many meters long should the race be in order for Dhvan to win the following calculation must be performed:

45 / (2.5 - 1) = X

45 / 1.5 = X

30 = X

Therefore, the race must be up to 29 meters for Dhvan to win.

6 0
3 years ago
A recent study done by the National Retail Federation found that 2019 back-to-school spending for all US households who have sch
MissTica

Answer:

Step-by-step explanation:

Hello!

The working variable is:

X: Back-to-school expense of a US household with school-aged children.

X~N(μ;σ²)

μ= $697

σ= $120

a. What is the probability that 2019 back-to-school spending for a US household with school-aged children is greater than $893?

Symbolically: P(X>$893)

First, you standardize the probability using Z= (X-μ)/σ ~N(0;1)

P(X>$893)= P(Z>(893-697)/120)= P(Z>1.63)

To resolve this question you have to use the table of cumulative probabilities for the standard normal distribution. These tables accumulate probabilities from the left, symbolically P(Z≤Z₀), so to reach probabilities greater than a Z₀ value you have to subtract the cumulative probability until that value from the maximum probability value 1:

P(Z>1.63)= 1 - P(Z≤1.63)= 1 - 0.94845= 0.05155

b. Provide the Z-score corresponding to the 2019 back-to-school spending of $1,200, and the probability of 2019 back-to-school spending for a household with school-aged children is less than $1,200.

P(X<$1200) = P(Z<(1200-697)/120)= P(Z<4.19)= 1

According to the empirical rule of the normal distribution, 99% of the data is between μ ± 3σ. This, logically, applies to the standard normal distribution. Considering that the distribution's mean is zero and the standard deviation is one, then 99% of the probabilities under the standard normal distribution are within the Z values: -3 and 3, values below -3 will have a probability equal to zero and values above 3 will have probability equal to one.

c. Find Q3 (Third Quartile).

Q3 in the value that marks three-quarters of the distribution, in other words, it has 75% of the distribution below it and 25% above, symbolically:

P(Z≤c)=0.75

In this case, you have to look in the center of the right Z-table (positive) for the probability of 0.75 and then the margins to find the Z-score that belongs to that cumulative probability:

c= 0.674

Now you reverse the standardization to see what value of X belongs to the Q3:

c= (X-μ)/σ

X= (c*σ)+μ

X= (0.674*120)+697= $777.88

d. Find Q1 (First Quartile)

To resolve this you have to follow the same steps as in c., just that this time you'll look for the value that marks the first quarter of the distribution, symbolically:

P(Z≤d)= 0.25

In this case, since the probability is below 0.5 you have to look for the Z value in the left table (negative).

d= -0.674

d= (X-μ)/σ

X= (d*σ)+μ

X= (-0.674*120)+697= $616.12

e. What is the value of the IQR for the distribution of 2019 back-to-school spending for a US household with school-aged children?

IQR= Q3-Q1= $777.88 - $616.12= $161.76

f. Interpret the value of the IQR from question 2e within the context of the problem.

$161.76 represents the distance between 75% of the Back-to-school expense of a US household 25% of the Back-to-school expense of US households.

g. What is the proportion of 2019 back-to-school spending within 1.50 standard deviations of the mean?

"Within 1.50 standard deviations of the mean" can be symbolized as "μ ± 1.5σ" or "μ - 1.5σ≤ Z ≤μ + 1.5σ"

P(μ - 1.5σ≤ Z ≤μ + 1.5σ)

Since the mean is zero and the standard deviation is one:

P(-1.5 ≤ Z ≤ 1.5)= P(Z≤1.5) - P(Z≤-1.5)= 0.933 - 0.067= 0.866

h. What is the 2019 back-to-school spending amount such that only 3% of households with school-age children spend more than this amount?

The "top" 3% means that you are looking for a value of the variable that has above it 0.03 of probability and below it 0.97%, first you look for this value under the standard normal distribution and then you reverse the standardization to reach the corresponding value of the variable:

P(Z>h)= 0.03 ⇒ P(Z≤h)=0.97

h= 1.881

h= (X-μ)/σ

X= (h*σ)+μ

X= ( 1.881*120)+697= $922.72

i. Which US household is more unusual, a US household with back-to-school spending of $600 or a US household with back-to-school spending of $900?

Under this kind of distribution, the "most usual" values are around the center (near the mean) and the "unusual" values will find themselves in the tails of the Gaussian bell.

To check which one is more unusual you have to see their distance with respect to the mean.

(X-μ)/σ

(600-697)/120= -0.8083

(900-697)/120= 1.69

An expense of $900 is more unusual than an expense of $600 (600 is almost the expected expenses)

j. Let's say the Smith family spent $815 on buying school supplies this fall. Provide an interpretation of the Smith family's 2019 back-to-school spending, i.e. what can you say about the percentage of all other US households with school-age children that have higher back-to-school spending than the Smith family?

P(X>$815) = P(Z>(815-697)/120)= P(Z>0.98)

1-P(Z≤0.983)= 0.837

83.7% of the families will have back-to-school expenses of $815 or more.

I hope it helps!

6 0
3 years ago
Someone please help!!! This is due tonight!! I’ll give brainliest
3241004551 [841]

Answer:

($0.5 x 2) + ($1.25 x 3) + ($0.75 x 3) = $7

6 0
3 years ago
Read 2 more answers
Can you help me pleeeeaaassee??
melamori03 [73]

Error 1: DT / TS × CT / TR which is the first error. We can fix the error by writing the correct equation RT / TD = ST / TC, Error 2: The second error is 7 / 16 × (x - 1) / 14 and we can fix the error by writing the equation 14 / 7 = 16 / (x - 1), Error 3:The third error is the value of x and we can find the correct value of x from the equation 14 / 7 = 16 / (x - 1) and the value of x is 9.

Given: The diagram is given and we need to find the errors and then fix them. Also ΔTSR ≈ ΔTCD

Let's solve the given question:

Given that ΔTSR ≈ ΔTCD

So we know by the properties of the similarity that if two triangles are similar then the ratio of their corresponding sides is equal.

So, ΔTSR ≈ ΔTCD

=> RT / TD = ST / TC

=> 14 / 7 = 16 / (x - 1)

In the question, we can observe that the given side ratio is DT / TS × CT / TR which is the first error. We can fix the error by writing the correct equation RT / TD = ST / TC.

The second error is 7 / 16 × (x - 1) / 14 and we can fix the error by writing the equation 14 / 7 = 16 / (x - 1).

The third error is the value of x.

We can find the correct value of x from the given equation:

14 / 7 = 16 / (x - 1)

=> 2 = 16 / (x - 1)

Multiplying both sides by (x - 1):

(x - 1) × 2 = 16 / (x - 1) × (x - 1)

=> 2(x - 1) = 16

Multiplying both sides by 1 / 2:

2(x - 1) × 1 / 2= 16 × 1 / 2

=> x - 1 = 8

Adding 1 on both sides:

x - 1 + 1 = 8 + 1

x = 9

Therefore x = 9.

Hence the errors are:

Error 1: DT / TS × CT / TR which is the first error. We can fix the error by writing the correct equation RT / TD = ST / TC

Error 2: The second error is 7 / 16 × (x - 1) / 14 and we can fix the error by writing the equation 14 / 7 = 16 / (x - 1).

Error 3:The third error is the value of x and we can find the correct value of x from the equation 14 / 7 = 16 / (x - 1) and the value of x is 9.

Know more about "similar triangles" here: brainly.com/question/14366937

#SPJ9

8 0
2 years ago
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