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3 years ago

Mr. Thomas drove 75 miles in May. He drove 6 times as many miles in July as he did in May. He

Mathematics

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

1800 miles

Step-by-step explanation:

No. of miles driven by Mr. Thomas in May = 75

It is given that miles driven in July is 6 times of miles driven by Mr. Thomas in May(75 miles).

Thus

No. of miles driven by Mr. Thomas in July = 6 * No. of miles driven by Mr. Thomas in May = 6*75 = 450 miles.

__________________________________________________

Another condition given that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).

Thus

No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July = 4* 450 miles = 1800 miles.

No. of miles driven by Mr. Thomas in June is 1800 miles.

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Nastasia [14]

Answer:

Step-by-step explanation:

You are correct

4 0

3 years ago

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

4 0

3 years ago

Ricole wrote a number that is the same distance from 0 to 15 on a number line but in a opposite direction . What number did rico

pshichka [43]

Answer:

1 2 3 4 5 6 7 8 9 10 11 12 12 14 15!

5 0

3 years ago

Tyler had $65.40 in his checking account. Then he wrote checks on the account for $20.38, $11.48, and $19.50. What is the balanc

andrew-mc [135]

money going out (20.38+11.48+19.50) =51.36

balance - out = new balance

65.40-51.36 =

14.04

new balance = 14.04

4 0

3 years ago

A 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense he

Orlov [11]

Answer:

We conclude that the the touch therapists does not use a method equivalent to random guesses.

Yes, the results suggest that touch therapists are effective.

Step-by-step explanation:

We are given that in a 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense her energy field.

Among 264264 trials, the touch therapists were correct 105105 times.

Let p = proportion that touch therapists uses a random guess method.

Here, random guess means; p = 50%

So, Null Hypothesis, $H_0$ : p = 50% {means that the touch therapists use a method equivalent to random guesses}

Alternate Hypothesis, $H_A$ : p $\neq$ 50% {means that the touch therapists does not use a method equivalent to random guesses}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of touch therapists were correct = $\frac{105}{264}$ = 0.39

n = sample of trials = 264

So, test statistics = $\frac{\frac{105}{264} -0.50}{\sqrt{\frac{\frac{105}{264}(1-\frac{105}{264})}{264} } }$

= -3.395

The value of z test statistics is -3.395.

Now, at 0.05 significance level the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the the touch therapists does not use a method equivalent to random guesses.

Yes, the results suggest that touch therapists are effective.

3 0

2 years ago