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2 years ago

Ram travels 200 km in 2 and a half hours by a car, how long would he take to travel 140

Mathematics

1 answer:

MrRissso [65]2 years ago

4 0

Answer:

x=1.75 hour

Step-by-step explanation:

200km=2.5

140km=x

2.5 * 140=200x

350=200x

x=350/200

x=1.75

the time it will take is x=1.75 hour

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Which number completes this Pythagorean triple 300, 400 ?

forsale [732]

300²+400²=500² is your answer
a² + b² = c²

8 0

3 years ago

A hiker starting at point P on a straight road wants to reach a forest cabin that is 2 km from a point Q, 3 km down the road fro

pantera1 [17]

Answer:

2.19 km

Step-by-step explanation:

If x is the distance she walks down the road before turning, then the total time is:

t = x/8 + √((3 − x)² + 2²) / 3

t = x/8 + √(9 − 6x + x² + 4) / 3

24t = 3x + 8√(13 − 6x + x²)

24t = 3x + 8(13 − 6x + x²)^½

Take derivative of both sides with respect to x.

24 dt/dx = 3 + 4(13 − 6x + x²)^-½ (-6 + 2x)

When t is a minimum, dt/dx = 0.

0 = 3 + 4(13 − 6x + x²)^-½ (-6 + 2x)

-3 = 4(13 − 6x + x²)^-½ (-6 + 2x)

3 / (6 − 2x) = 4(13 − 6x + x²)^-½

3 / (24 − 8x) = (13 − 6x + x²)^-½

(24 − 8x) / 3 = (13 − 6x + x²)^½

(24 − 8x)² / 9 = 13 − 6x + x²

576 − 384x + 64x² = 117 − 54x + 9x²

459 − 330x + 55x² = 0

Solve with quadratic formula.

x = [ 330 ± √((-330)² − 4(55)(459)) ] / 2(55)

x = (330 ± √7920) / 110

x = 2.19 or 3.81

Since 0 < x < 3, x = 2.19.

7 0

3 years ago

What is the interquartile range (IQR) of the following data set 17 16 21 15 25 22 18 23 17

Pepsi [2]

IQR = 6

First locate the median $Q_{2}$ at the centre of the data arranged in ascending order. Then locate the lower and upper quartiles $Q_{1}$ and $Q_{3}$ located at the centre of the data to the left and right of the median.

Note that if any of the above are not whole values then they are the average of the values either side of the centre.

rearrange data in ascending order

15 16 ↓17 17 18 21 22 ↓23 25

↑

$Q_{2}$ = 18

$Q_{1}$ = $\frac{16+17}{2}$ = 16.5

$Q_{3}$ = $\frac{22+23}{2}$ = 22.5

IQR = $Q_{3}$ - $Q_{1}$ = 22.5 - 16.5 = 6

4 0

3 years ago

Helpppppp mathhh failing

LenKa [72]

Sin60 =b/10
√3/2 = b/10
b = √3 / 2 (10)
b = 5√3

d^2 = 10^2 - (5√3)^2
d^2 = 100 - 75
d^2 = 25
d = 5

sin30 = b/a
1/2 = 5√3 / a

a = 5√3 (2)
a = 10√3

c^2 = (10√3)^2 - (5√3)^2
c^2 = 300 - 75
c^2 = 225
c = 15

so
a =10√3, b = 5√3, c = 15,d = 5
answer is
D. last choice

4 0

3 years ago

The Genetics & IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceivi

Reika [66]

Answer: The test statistic needed to test this claim= 10.92

Step-by-step explanation:

We know that the probability of giving birth to a boy : p= 0.5

i..e The population proportion of giving birth to a boy = 0.5

As per given , we have

Null hypothesis : $H_0: p\leq0.5$

Alternative hypothesis : $H_a: p>0.5$

Since $H_a$ is right-tailed , so the hypothesis test is a right-tailed z-test.

Also, it is given that , the sample size : n= 291

Sample proportion: $\hat{p}=\dfrac{239}{291}\approx0.82$

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$ , where n is sample size , $\hat{p}$ is sample proportion and p is the population proportion.

$\Rightarrow\ z=\dfrac{0.82-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{291}}}\approx10.92$

i.e. the test statistic needed to test this claim= 10.92

Critical value ( one-tailed) for 0.01 significance level = $z_{0.01}=2.326$

Decision : Since Test statistic value (10.92)> Critical value (2.326), so we reject the null hypothesis .

[When test statistic value is greater than the critical value , then we reject the null hypothesis.]

Thus , we concluded that we have enough evidence at 0.01 significance level to support the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy.

7 0

3 years ago