PLEASE HELP!!

Rachel runs 2km to her bus stop, and then rides 4.5 km to school. On average, the bus is 45 km/h faster than Rachel's average ru

Mice21 [21]

Answer:6.042km/h 6km/h approximately

Step-by-step explanation:

First off we have to know the formula relating speed, distance and time which is

Speed = distance/time

Now we are looking for Rachel's running speed

We are to find Rachel's running speed, so let's label is x

We are given that the distance Rachel runs to her bus stop is 2km

We were not given the time she uses to run to the bus stop

So let's label the time Rachel uses to run to her bus stop as y

So from the formula speed = distance/time

We have x = 2/y

Now we are told that the speed the bus uses to get to school is 45km/h faster than her speed used to run

So speed of bus = 45 + x

And the overall time for the whole journey is 25mins, changing this to hours, because the speed details given is in km/h we divide 25 by 60 which will give 0.417

Now if the total time is 0.417 hours, and we labeled the time for Rachel to run to the bus as y, so the time for the time for the bus to get to school will be 0.417 - y

We are also told the bus rides for 4.5km to school

So adding this together to relate the speed, distance and time of the bus with the formula speed = distance/time

We get 45 + x = 4.5/(0.417 - y)

So we have two equations

x = 2/y (1)

45+x = 4.5/(0.417-y) (2)

So putting (1) in (2) we have

45 + (2/y) = 4.5/(0.417-y)

Expanding further

(45y + 2)/y = 4.5(0.417-y)

Cross multiplying

(45y + 2)(0.417 - y) = 4.5y

Opening the brackets

18.765y - 45y2 + 0.834 - 2y = 4.5y

Collecting like terms

-45y2 + 18.765y -2y - 4.5y + 0.834 = 0

-45y2 + 12.265 + 0.834 = 0

Dividing all sides by -45 to make the coefficient of y2 1

y2 - 0.273y - 0.019 = 0

Now we have gotten a quadratic equation, and since it's with decimal numbers we can use either completing the square method of almighty formula

I'm using almighty formula her

For solving

ax2 + bx + c = 0

x = (-b +-root(b2-4ac)/2a

For our own equation, we are finding y

From our our quadratic equation

a = 1, b=-0.273, c = -0.834

you = (-(-0.273)+-root(-0.273-4(1)(-0.019))/2(1)

y = (273+-root(0.151))/2

y = (0.273+0.389)/2 or (0.273-0.389)/2

y = 0.331 or -0.085

So we use the positive answer which is 0.331, because time can't be negative

Then we put y = 0.331 in (1)

x = 2/y

x = 2/0.331

x = 6.042km/h

x = 6km/h approximately

4 0

2 years ago

Please help me will mark brainliest!!! Provide a hand drawn graph attachment pls

Travka [436]

The graph is shown below. I used GeoGebra to create the graph. The graph is restricted on the domain $0 \le x \le 2$ meaning that everything to the left of x = 0 is not drawn, and the same for everything to the right of x = 2.

A table of values is included as well. Each row in the table represents an ordered pair point (x,y) that is on the blue cosine graph.

5 0

2 years ago

3. 100x2-36y2 ?? ito pa guys pleasedddd!!

uranmaximum [27]

Answer:

4(5x - 3y)(5x + 3y)

Step-by-step explanation:

Assuming you require the expression factored.

Given

100x² - 36y² ← factor out 4 from each term

= 4(25x² - 9y²) ← is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , thus

25x² - 9y²

= (5x)² - (3y)² ← with a = 5x and b = 3y

= (5x - 3y)(5x + 3y)

Thus

100x² - 36y² = 4(5x - 3y)(5x + 3y) ← in factored form

8 0

2 years ago

Solve for x x/12−1=−6 x=

Alik [6]

Answer:

X=? Sorry

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The most recent public health statistics available indicate that 23.6% of American adults smoke cigarettes. Using the 68-95-99

artcher [175]

Answer:

There is a 68% chance that between 17% and 30% are smokers.

There is a 95% chance that between 10% and 37% are smokers.

There is a 99.7% chance that between 4% and 44% are smokers.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of the sampling distribution of sample proportion is:

$\mu_{\hat p}=p\\$

The standard deviation of the sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

Given:

n = 40

p = 0.236

Compute the mean and standard deviation of this sampling distribution of sample proportion as follows:

$\mu_{\hat p}=p=0.236$

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.236(1-0.236)}{40}}=0.067$

The Empirical Rule states that in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be divided into three parts:

68% data falls within 1 standard-deviation of the mean.

That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.

95% data falls within 2 standard-deviations of the mean.

That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.

99.7% data falls within 3 standard-deviations of the mean.

That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

Compute the range of values that has a probability of 68% as follows:

$P (\mu_{\hat p} - \sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + \sigma_{\hat p}) = 0.68\\P(0.236-0.067\leq \hat p \leq 0.236+0.067)=0.68\\P(0.169\leq \hat p \leq0.303)=0.68\\P(0.17\leq \hat p \leq0.30)=0.68$

Thus, there is a 68% chance that between 17% and 30% are smokers.

Compute the range of values that has a probability of 95% as follows:

$P (\mu_{\hat p} - 2\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 2\sigma_{\hat p}) = 0.95\\P(0.236-2\times 0.067\leq \hat p \leq 0.236+2\times0.067)=0.95\\P(0.102\leq \hat p \leq 0.370)=0.95\\P(0.10\leq \hat p \leq0.37)=0.95$

Thus, there is a 95% chance that between 10% and 37% are smokers.

Compute the range of values that has a probability of 99.7% as follows:

$P (\mu_{\hat p} - 3\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 3\sigma_{\hat p}) = 0.997\\P(0.236-3\times 0.067\leq \hat p \leq 0.236+3\times0.067)=0.997\\P(0.035\leq \hat p \leq 0.437)=0.997\\P(0.04\leq \hat p \leq0.44)=0.997$

Thus, there is a 99.7% chance that between 4% and 44% are smokers.

7 0

2 years ago