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3 years ago

Can anyone please answer question 1 please it’s for my homework!! Xx

Mathematics

1 answer:

astraxan [27]3 years ago

6 0

1/4 of her original money is 1.25 + 1.60 = 2.85

She started with 2.85 * 4 = 11.40 pounds.

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What is 965,000,000,000,000 in scientific notation?

stellarik [79]

The answer is 9.65x10^14 because in order for the decimal point to be between 9 and 6 you need to move 14 decimal places back

4 0

3 years ago

Answer ASAP, please..

Phantasy [73]

Answer:

g(x) = $(x-3)^{2} -4$

Step-by-step explanation:

The graph is shifted 3 units to the right which is represented by -3 (in the parentheses) and is shifted four units down.

6 0

3 years ago

You rent an apartment that costs

Inessa [10]

Answer:

$2,660

Step-by-step explanation:

9.5% of 1300 is 123.5

So to figure out what the rent would cost in 11 years, we just have to multiply 123.5 by 11.

123.5 × 11 = 1358.5

We can then add that number to 1300.

1358.5 + 1300 = $2658.5

Then we round.

Your final answer would be:

$2,660

6 0

2 years ago

Read 2 more answers

Find the value of x and y<br><img src="https://tex.z-dn.net/?f=2x%20%2B%205y%20%3D%20%20-%205%20%5C%5C%20%20-%208x%20-%201y%20%3

Gemiola [76]

Answer:

(x,y)=(0,-1)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago