If I drive 70 miles per hour for eight and a half hours, how far will I have driven

Stretch thinking Rewrite this sentence using the word fewer. Carey reads 10more pages than Lucey.

anzhelika [568]

Answer:

Lucey reads 10 pages fewer than Carey.

4 0

3 years ago

What statement about a rhombus is never true?

Alekssandra [29.7K]

It can never be a square.

7 0

3 years ago

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An electrician needs to cut a 49 m piece of wire into two pieces so that one piece is 5 m longer than the other piece. What is t

gulaghasi [49]

The length of the shorter wire that the electrician cuts so that the difference between those two is of 5 m is 22.5.

Given to us

An electrician needs to cut a 49 m piece of wire into two pieces so that one piece is 5 m longer than the other piece.

<h3>What are the two equal parts of the wire?</h3>

We know that the length of the wire is 49 m, therefore, to cut the wire in two equal parts divide the total length of the wire by two,

$\dfrac{49}{2} = 24.5$

Thus, the length of each side will be 24.5.

<h3>What is the length of the shorter piece of wire?</h3>

Since we need to cut the wire such that one side of the wire is 5 less than the other, we already know the length of the equal sides deduct (5/2) from both sides to create a difference between the length of the wire.

Length of the shorter wire = 24.5 -2.5 = 22.5

Length of the longer wire = 22.5 + 5 = 27.5

Hence, the length of the shorter wire that the electrician cuts so that the difference between those two is of 5 m is 22.5.

Learn more about Lengths:

brainly.com/question/8552546

3 0

2 years ago

1. Average Senior High School annual cost of tuition fee for all private schools in Cavite last year was 46,300 a random sample

anygoal [31]

Answer:

Step-by-step explanation:

H0 : μ = 46300

H1 : μ > 46300

α = 0.05

df = n - 1 = 45 - 1 = 44

Critical value for a one tailed t-test(since population standard deviation is not given).

Tcritical = 1.30

The test statistic :(xbar - μ) ÷ (s/sqrt(n))

The test statistic, t= (47800-46300) ÷ (5600√45)

t = 1500

t = 1500 / 834.79871

t = 1.797

The decision region :

Reject H0: if t value > critical value

1. 797 > 1.30

Tvalue > critical value ; We reject H0

Hence, there is sufficient evidence to conclude that cost has increased.

5 0

3 years ago

A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili

Fofino [41]

Answer:

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

3 defects - 18 square feet

x defects - 31 square feet

$18x = 3*31$

$x = \frac{31*3}{18}$

$x = 5.17$

So $\mu = 5.17$

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

So

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

So

$P(X \geq 4) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057$

$P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294$

$P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760$

$P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309$

So

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242$

Finally

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758$

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

8 0

3 years ago