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PilotLPTM [1.2K]

3 years ago

12

a teacher needs 70 lengths of string cut to 40 cm each. If balls of string are 10 m long how many balls will be needed

1 answer:

LenKa [72]3 years ago

7 0

A meter is 100 cm. A ball of string has 10m of string, and you multiply 10m by 100cm, you get 1,000cm. Then you multiply 40cm by 2, and you get 80cm, or two lengths of string. Then you multiply 35, which is half of 70, by 80, and that is 2,800 cm.
2,800÷40cm=70
And 100cm=1m.
The teacher will therefore need 28 balls of string.

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3 years ago

A biased dice is thrown.

nexus9112 [7]

Given:

A biased dice is thrown 300 times.

Table of probabilities of each score.

To find:

The expected number of times the score will be odd.

Solution:

Odd numbers on the dice are 1, 3, 5. The sum of their probability is

$0.15+0.25+0.3=0.7$

Even numbers on the dice are 2, 4, 6. The sum of their probability is

$0.05+0.05+0.2=0.3$

Now, the expected number of times the score will be odd is

$\text{Expected odd number}=300\times \text{Probability of getting an odd number}$

$\text{Expected odd number}=300\times 0.7$

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3 years ago

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Olin [163]

Answer:

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Step-by-step explanation:

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3 years ago

Simplify (w^6c) (-3w^4c^3)

yulyashka [42]

<span>(w^6c) (-3w^4c^3)
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3 years ago

In a recent year the fuel economy of all passenger vehicles was 19.8 mpg. A trade organization sampled 50 passenger vehicles for

salantis [7]

Answers and Step-by-step explanation:

Hypotheses are:

H o=19.8

H 1 ≠19.8

Here we have following information:

s=2.45, X bar=20.1, n=50

So test statistics will be

t=20.1-19.8/(2.45/ $\sqrt{50}$ =0.87

Degree of freedom : d f = n - 1 = 50 - 1 =49

Test is two tailed so critical value of the test is -1.30 and 1.30.

Rejection region:

if t < -1.30 and t > 1.30, reject H o

Since test statistics lie in the rejection region so we fail to reject the null hypothesis.

(b)

The observed significance of the test is

P value =0.3885

(c)

Since p value is greater than 0.20 so we fail to reject the null hypothesis.

5 0

3 years ago