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ch4aika [34]
3 years ago
15

Find the discriminantx^2-x+5=0​

Mathematics
1 answer:
Feliz [49]3 years ago
6 0

Answer:

20

Step-by-step explanation:

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Find the midpoint of the line segment whose endpoints are (3, 10) and (7, 8).
Stels [109]

Answer:

Option 2 is correct.

Step-by-step explanation:

Given the coordinates of lines segment (3, 10) and (7, 8). we have to find the mid-point of given line segment.

Mid-point formula states that if (x_1,y_1) and (x_2,y_2) are the coordinates of end points of line segment then the coordinates of mid-point are

(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

∴ Coordinates of mid-point of line segment joining the points          (3, 10) and (7, 8) are

(\frac{3+7}{2},\frac{10+8}{2})=(\frac{10}{2},\frac{18}{2})=(5,9)

Hence, option 2 is correct.

6 0
2 years ago
Read 2 more answers
Assume that y varies inversely with x. If y = 7 when x = 2/3, find y when x = 7/3. <br>y = [?] ​
Alekssandra [29.7K]

Answer:

y = 2

Step-by-step explanation:

varies inversely

xy = k

y = 7 when x = 2/3

(2/3)7 = k

14/3 = k

k = 14/3

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

find y when x = 7/3

(7/3)y = 14/3

multiply both sides by 3/7

y = 14/3 * 3/7

y = 2

4 0
2 years ago
Read 2 more answers
Find midpoint of CD C=(10,-1)and D=(6,3)
antoniya [11.8K]

Answer:

(8,1)

Step-by-step explanation:

The x coordinate of the midpoint is

(10+6)/2 = 16/2  =8

The y coordinate  of the midpoint is

(-1+3)/2 = 2/2 =1

The midpoint is (8,1)

6 0
3 years ago
15 billion in numerals?
Leokris [45]
A billion has 9 zeros. So
15,000,000,000
:)
4 0
3 years ago
Read 2 more answers
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
3 years ago
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