The number 15 has 2 place values.
The value in the tens place is 1.
The value in the ones place is 5.
1 tens + 5 ones = 15
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Answer:
Sol: In this question, firstly we have to make the first bracket as a complete square of the second bracket. This we can by adding 2.x.1/x which is equivalent to 2. Then the equation becomes:
6(x2 + 1/x2 +2) – 5(x + 1/x) = 50 { 38 + 6*2)
⇒ 6(x2 + 1/x2 +2) – 5(x + 1/x) – 50 = 0
Now put x + 1/x = y
⇒ 6y2 -5y -50 = 0
⇒ (2y +5)(3y-10)= 0
⇒ y=-5/2 or 10/3
As x is positive therefore, x + 1/x =10/3
On solving further you will get as x=3 or 1/3
Answer: B. (-6, 5)
Step-by-step explanation:
Answer:
Yes, we can assume that the percent of female athletes graduating from the University of Colorado is less than 67%.
Step-by-step explanation:
We need to find p-value first:
z statistic = (p⁻ - p0) / √[p0 x (1 - p0) / n]
p⁻ = X / n = 21 / 38 = 0.5526316
the alternate hypothesis states that p-value must be under the normal curve, i.e. the percent of female athletes graduating remains at 67%
H1: p < 0.67
z = (0.5526316 - 0.67) / √[0.67 x (1 - 0.67) / 38] = -0.1173684 / 0.076278575
z = -1.538681
using a p-value calculator for z = -1.538681, confidence level of 5%
p-value = .062024, not significant
Since p-value is not significant, we must reject the alternate hypothesis and retain the null hypothesis.
