Ask question

Ask question

All categories

2 years ago

12

Lisa is standing on a dock beside a lake. She drops a rock from her hand into the lake. After the rock hits the surface of the l

ake, the rock's distance from the lake's surface changes at a rate of -5 inches per second. If lisa holds her hand 5 feet above the lake's surface, how far from Lisa's hand is the rock 4 seconds after it hits the surface? Please show how you arrive at your answer

1 answer:

yawa3891 [41]2 years ago

5 0

Well, it goes 5 inches per second so after 4 seconds it is 4 times 5 =20 inches under the surface, add that to 5 feet and you get 6 ft and 8 inches

You might be interested in

Beatrice graphed the relationship between the

sdas [7]

Answer:

Option C. The time in seconds that passed before the printer started printing pages

see the explanation

Step-by-step explanation:

Let

y ---->the number of pages printed.

x ---> the time (in seconds) since she sent a print job to the printer

we know that

The x-intercept is the value of x when the value of y is equal to zero

In the context of the problem

The x-intercept is the time in seconds that passed before the printer started printing pages (the number of pages printed is equal to zero)

5 0

2 years ago

Read 2 more answers

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0

2 years ago

Which answer choice contains all the factors of 10

RSB [31]

Factors of 10 are : 1,2,5,10

7 0

2 years ago

A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly locat

DiKsa [7]

Answer:

The probability of a selection of 50 pages will contain no errors is 0.368

The probability that the selection of the random pages will contain at least two errors is 0.2644

Step-by-step explanation:

From the information given:

Let q represent the no of typographical errors.

Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let $\mu$ be the random variable that follows a Poisson distribution, then mean $\mu = \dfrac{10}{500}= 0.02$

and the mean that the random selection of 50 pages will contain no error is $\lambda = 50 \times 0.02 =1$

∴

$Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}$

Pr(q =0) = 0.368

The probability of a selection of 50 pages will contain no errors is 0.368

The probability that 50 randomly page contains at least 2 errors is computed as follows:

P(X ≥ 2) = 1 - P( X < 2)

P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2

$P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]$

$P(X \geq 2) = 1 - [0.3678 +0.3678]$

$P(X \geq 2) = 1 -0.7356$

P(X ≥ 2) = 0.2644

The probability that the selection of the random pages will contain at least two errors is 0.2644

6 0

2 years ago

for a school fundraiser troy sold 28 bags of popcorn and 40 candy bars and made 282. jake sold 17 bags of popcorn and 20 candy b

Nataliya [291]

Each bag of popcorn is equal to 6.5 dollars

6 0

2 years ago