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4 years ago

If James can swim 10 laps in 23 minutes, what would be his expected time for 15 laps.(show work please)

Mathematics

2 answers:

Nana76 [90]4 years ago

4 0

Solution:

we have been given that

James can swim 10 laps in 23 minutes.

we have been asked to find the expected time for 15 laps.

Since James can swim 10 laps in 23 minutes.

It mean that James can swim 1 lap in $=\frac{23}{10}=2.3 \ minutes$

So for 15 laps expected time will be $=\frac{23}{10}*15= 34.5 minutes$

Delicious77 [7]4 years ago

3 0

10/23=15/m
(15x23)/10
343/10
34.3

so....
10 laps in 23 minutes
=
15 laps in 34.3 minutes

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3 years ago

write a rational expression involving one variable for which the excluded values are -4 and -7. Please Help!!!!!!

lions [1.4K]

Answer:

See below.

Step-by-step explanation:

Values are excluded if they make the denominator zero, so one rational expression could be:

(x^2 - 9) / (x + 4)(x + 7)

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3 years ago

Find the magnitude of WX for W(-2, 8, -3) and X(1, 4, -1)

zheka24 [161]

The magnitude of WX is $\sqrt{29}$ ⇒ c

Step-by-step explanation:

The magnitude of XY (its length) where,

$X=(x_{1},y_{1},z_{1})$
$Y=(x_{2},y_{2},z_{2})$

is IXYI = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

∵ W = (-2 , 8 , -3)

∵ X = (1 , 4 , -1)

- To find the magnitude of WX use the formula above

∴ $x_{1}$ = -2 and $x_{2}$ = 1

∴ $y_{1}$ = 8 and $y_{2}$ = 4

∴ $z_{1}$ = -3 and $z_{2]$ = -1

- Substitute these values in the rule above

∵ IWXI = $\sqrt{(1--2)^{2}+(4-8)^{2}+(-1--3)^{2}}=\sqrt{9+16+4}$

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3 years ago

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3 years ago

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The circumference of a sphere was measured to be 82 cm with a possible error of 0.5 cm.

Nat2105 [25]

Answer:

A) Maximum error = 170.32 cm³

B)Relative error = 0.0575

Step-by-step explanation:

A) Formula for circumference is: C = 2πr

Differentiating with respect to r, we have;

dC/dr = 2π

r is small, so we can write;

ΔC/Δr = 2π

So, Δr = ΔC/2π

We are told that ΔC = 0.5.

Thus; Δr = 0.5/2π = 0.25/π

Now, formula for Volume of a sphere is;

V(r) = (4/3)πr³

Differentiating with respect to r, we have;

dV/dr = 4πr²

Again, r is small, so we can write;

ΔS/Δr = 4πr²

ΔV = 4πr² × Δr

Rewriting, we have;

ΔV = ((2πr)²/π) × Δr

Since C = 2πr, we now have;

ΔV = (C²/π)Δr

ΔV will be maximum when Δr is maximum

Thus, ΔV = (C²/π) × 0.25/π

C = 82 cm

Thus;

ΔV = (82²/π) × 0.25/π

ΔV = 170.32 cm³

B) Formula for relative error = ΔV/V

Relative error = 170.32/((4/3)πr³)

Relative error = 170.32/((4/3)C³/8π³)

Relative errror = 170.32/((4/3)82³/8π³)

Relative error = 170.32/2963.744

Relative error = 0.0575

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3 years ago