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3 years ago

I have no clue what I’m doing can someone help me with number 5

Mathematics

1 answer:

marin [14]3 years ago

5 0

a) 35w+55-10w = 25w+55

28w+75-5w+12 = 23w +87

b) 25w+55 + 23w +87

=48w+142

they will be short by 2(24w+71)

c) 50w+25 = 25(2w+1)

answer: =98w+167

i think

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Answer:

A. Subtract 6 to both sides.

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Answer:

$P(A\ and\ B) = \frac{3}{7}$

Step-by-step explanation:

Your question is not well presented; (See Attachment for complete details)

Required

Find P(A n B)

where

A = Event that the place is a city

B = Event that the place is in North

The first step is to get the sample space;

Let S represent the sample space;

$S = \{India,\ Tokyo,\ Houston,\ Peru,\ New York,\ Tijuana,\ Canada \}$

$n(S) = 7$

The next is to list events A and B

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$A = \{Tokyo,\ Houston,\ New York,\ Tijuana\}$

B = North America

$B = \{Houston,\ New York,\ Tijuana,\ Canada\}$

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$A\ n\ B = \{Houston,\ New York,\ Tijuana\}$

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The probability of A and B is calculated as follows;

$P(A\ and\ B) = \frac{n(A\ and\ B)}{n(S)}$

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$P(A\ and\ B) = \frac{3}{7}$

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4 years ago

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Answer:

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Step-by-step explanation:

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A county is considering rasing the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 m

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Answer:

No, at alpha equals 0.10, we do not have enough evidence to support the county's claim.

Step-by-step explanation:

We are given that a county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 miles per hour.

A random sample of 15 vehicles has a mean speed of 31 miles per hour and a standard deviation of 4.7 miles per hour.

Let $\mu$ = true mean speed of the vehicles.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 30 miles per hour {means that the mean speed of vehicles is lesser than or equal to 30 miles per hour}

Alternate Hypothesis, $H_A$ : $\mu$ > 30 miles per hour {means that the mean speed of vehicles is greater than 30 miles per hour}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean speed of 15 vehicles = 31 mph

s = sample standard deviation = 4.7 mph

n = sample of vehicles = 15

So, test statistics = $\frac{31-30}{\frac{4.7}{\sqrt{15} } }$ ~ $t_1_4$

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Hence, the value of test statistics is 0.824.

Now at 0.10 significance level, the t table gives critical value of 1.345 at 14 degree of freedom for right-tailed test. Since our test statistics is less than the critical value of t as 0.824 < 1.345, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean speed of vehicles is lesser than or equal to 30 miles per hour which means that the county's claim is not supported.

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