Answer: the IQR is 3.5
Step-by-step explanation:
Answer:
0.0016
Step-by-step explanation:
Batting average, p = 0.26
n = 7
x = 6
With p = 0.26 as success rate
1-p is equal to failure rate which is = 0.74
We have to solve this by using the binomial distribution formula.
P(X= x)
= nCx * p^x * (1-p)^(n-x)
P(X = 6)
=7C6 × 0.26^6 ×(1-0.26)^(7-6)
= 7 × 0.0003089 × 0..74¹
= 0.0016
So probability that he has exactly 6 hits in his next 7 bats is equal to 0.0016.
Answer:
2x + 5y = -5
Step-by-step explanation:
Since the new line is parallel to the given line 2x + 5y = 10, the equation of the new line has exactly the same form as does 2x + 5y = 10, except that the constant will be different.
Were we to solve this equation (in standard form) for y in slope-intercept form, we'd get:
5y = -2x + 10, or
-2x + 10
y = ----------------
5
or
y = (-2/5)x + 2
Writing out 2x + 5y = C, we substitute -5 for x and 1 for y, obtaining
2x + 5y = C => 2(-5) + 5(1) = -10 + 5 = -5. Therefore, C = -5 and the equation of the new line is
2x + 5y = -5
Answer:you might need up to 170 points toincrease the averge Points by 3
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A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
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A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
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prev Statement of a problem № 2618 next
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
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