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sergiy2304 [10]

3 years ago

12

At the beginning of a snowstorm, Camden had 10 inches of snow on his lawn. The snow then began to fall at a constant rate of 2.5

inches per hour. Assuming no snow was melting, how much snow would Camden have on his lawn 4 hours after snow began to fall? How much snow would Camden have on his lawn after t hours of snow falling?

1 answer:

Setler [38]3 years ago

8 0

Answer:

He would have a total of 20 inches of snow on his lawn

Step-by-step explanation:

2.5 x 4 = 10

10 + 10 = 20

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A sports stadium has 10000 seats, divided into box seats, lower-deck seats, and upper-devk seats. Box seats sell for 10 dollars,

lesantik [10]

Answer:

Number of box seats = 1000

Number of lower deck seats = 5000

Number of upper deck seats = 4000

Step-by-step explanation:

Let number of box seats = $x$

Let number of lower deck seats = $y$

Let number of upper deck seats = $z$

As per question statement:

$x+y+z=10000$ ...... (1)

Cost for box seat = $10

Cost for $x$ box seats = 10 $x$

Cost for lower deck seat = $8

Cost for $y$ lower deck seats = 8 $y$

Cost for lower deck seat = $5

Cost for $z$ lower deck seats = 5 $z$

As per question statement:

$10x+8y+5z=70000$ ..... (2)

$z = 4x$ ..... (3)

Putting value of $y$ in (1) and (2):

$x+y+4x=10000\\\Rightarrow 5x+y=10000$ ....... (4)

$10x+8 y+5\times 4x=70000\\\Rightarrow 30x+8y=70000 ..... (5)$

8 $\times$ (4) - (5):

$10x =$ 10000

$\Rightarrow x = 1000$

By (3):

$z = 4000$

Now, by equation (1):

$y$ = 5000

Number of box seats = 1000

Number of lower deck seats = 5000

Number of upper deck seats = 4000

6 0

3 years ago

1.Find the compound interest on Rs25000 for 3 years at 10% per annum ,Compounded annually.

LekaFEV [45]

Answer:

The compound interest is Rs8275

Step-by-step explanation:

The rule of the compound interest is A = P $(1+\frac{r}{n})^{nr}$ , where

A is the new amount

P is the initial amount

r is the interest rate in decimal

n is the number of periods

t is the time

The interest I = A - P

∵ The amount of investment is Rs25000 for 3 years

∴ P = 25000

∴ t = 3

∵ The rate of interest is 10% per annum, compounded annually

∴ r = 10% = 10 ÷ 100 = 0.1

∴ n = 1 ⇒ compounded annually

→ Substitute these value in the 1st rule above to find the new amount

∵ A = 25000 $(1+\frac{0.1}{1})^{1(3)}$

∴ A = 25000 $(1.1)^{3}$

∴ A = Rs33275

→ Use the 2nd rule above to find the interest

∵ I = 33275 - 25000

∴ I = 8275

∴ The compound interest is Rs8275

6 0

2 years ago

4.9cm cubed to mm cubed. Help me asap. google is a liar

Lunna [17]

It is 4900mm^3 because there are 1000mm^3 in one 1cm^3.
If you do 10^3 (10 mm) it gives you 1000mm^3.

Hope this helps.

8 0

3 years ago

Please help!!!Its timed !!!!! Ill give out most brainly and extra points !!!

seropon [69]

Answer:

<h2>D)x=4,y=7</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

equation
PEMDAS

<h3>let's solve:</h3>

according to the question

$\begin{cases}x + 29 = 7x + 5 \\ 8y - 35 =3y \end{cases}$

$\begin{cases}x -7x= 5-29 \\ 8y - 3y=35 \end{cases}$

$\begin{cases}-6x= -24\\ 5y=35 \end{cases}$

$\begin{cases}\frac{-6x}{-6}=\frac{-24}{-6} \\ \frac{5y}{5}=\frac{35}{5} \end{cases}$

$\therefore \begin{cases} x=4\\ y=7 \end{cases}$

6 0

3 years ago

Media experts claim that daily print newspapers are declining because of Internet access. Listed below, from left to right and

yKpoI14uk [10]

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic $t_{\alpha/2}$ is 6.678155

(d) The p value is 1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean, $\bar{x}_{1}$ = 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean, $\bar{x}_{2}}$ = 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The formula for t test is given by;

$t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic $t_{\alpha/2}$ = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.

5 0

3 years ago