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2 years ago

What is the distance between the points (2, -5) and (-5, -5) on a coordinate plane?

Mathematics

2 answers:

FromTheMoon [43]2 years ago

6 0

Answer:

Step-by-step explanation:

Both points are on the horizontal line y = -5. The distance between them is the difference of their x-coordinates:

2 -(-5) = 2+5 = 7

The distance between the points is 7 units.

kati45 [8]2 years ago

4 0

Answer:

The distance between the points (2, -5) and (-5, -5) on a coordinate plane is 7 units.

Step-by-step explanation:

Distance Formula: $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d=\sqrt{(-5-2)^{2}+(-5-(-5))^{2}}\\d=\sqrt{(-7)^{2}+(0)^{2}}\\d=\sqrt{49}\\d=7$

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3 years ago

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Tom and Jerry's Pizza Parlor changes $8.00 for large cheese pizza. Each additional topping is $1.

Scorpion4ik [409]

Answer:

Step-by-step explanation:

The price per pizza is $8.00 with additional toppings priced at $1.00 each.

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4 years ago

.In the Star Wars franchise, Yoda stands at only 66 centimeters tall. Suppose you want to see whether or not hobbits from the Lo

r-ruslan [8.4K]

Answer:

We conclude that the average height of hobbit is taller than Yoda.

Step-by-step explanation:

We are given that in the Star Wars franchise, Yoda stands at only 66 centimetres tall.

From a sample of 7 hobbits, you ﬁnd their mean height $\bar X$ = 80 cm with standard deviation s = 10.8 cm.

Let $\mu$ = average height of hobbit.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 66 cm {means that the average height of hobbit is shorter than or equal to Yoda}

Alternate Hypothesis, $H_A$ : $\mu$ > 66 cm {means that the average height of hobbit is taller than Yoda}

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

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

where, $\bar X$ = sample mean height = 80 cm

s = sample standard deviation = 10.8 cm

n = sample of hobbits = 7

So, test statistics = $\frac{80-66}{\frac{10.8}{\sqrt{7}}}$ ~ $t_6$

= 3.429

The value of t test statistics is 3.429.

Now, at 0.01 significance level the t table gives critical value of 3.143 for right-tailed test. Since our test statistics is more than the critical value of t as 3.429 > 3.143, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average height of hobbit is taller than Yoda.

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4 years ago

Bob spends 30 hours in 4 weeks of gardening. How many hours does he garden in 5 weeks

Mademuasel [1]

Answer:

37.5 hrs

Step-by-step explanation:

You do cross multiplication hours on top and weeks on the bottom.

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3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

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3 years ago