Which set(s) of numbers does − 5 4 − 4 5 belong to?

What value of Y will satisfy the given equation y divided by 2y — 15 = 7/5

guapka [62]

Answer:

$\huge\boxed{\sf y = \frac{41}{5} \ OR \ y = 8\frac{1}{5} }$

Step-by-step explanation:

$\sf 2y-15 = \frac{7}{5} \\\\Adding \ 15 \ to \ both \ sides\\\\2y = \frac{7}{5} + 15\\\\2y = \frac{7+75}{5} \\\\2y = \frac{82}{5} \\\\Dividing \ both \ sides \ bu 2\\\\y = \frac{82}{5*2} \\\\y = \frac{41}{5}$

Hope this helped!

<h2>~AnonymousHelper1807</h2>

3 0

3 years ago

My daughter needs help and I don't understand the question ethier.

WINSTONCH [101]

So firstly, let's figure out the expression for a singular goody bag. Since there's 2 of each item, a single goody bag is going to be $(1.07*2+0.85*2+2.12*2)$

Now since we have 7 friends who are all getting 1 goody bag, multiply the cost of the goody bag by 7, and our expression is $7(1.07*2+0.85*2+2.12*2)$ . Now we can solve it from there.

Firstly, solve the multiplication inside the parentheses: $7(2.14+1.7+4.24)$

Now add up what's inside the parentheses: $7(8.08)$

Lastly, multiply and your answer will be $56.56

3 0

3 years ago

Customers can pick their own apples at the happy apple farm. they pay $6 to enter the farm and $3.50 per pound for the apples th

Amanda [17]

Y = 3.5x + 6

Shows the 6 inital fee plus the 3.50 for each pound

4 0

3 years ago

Read 2 more answers

May you give me the answer to this please ?

Mrac [35]

Answer:

using pythagoras theorem

H^2=P^2+B^2

here h=MN ,opposite(P) be 15 and adjacent(B) be 16

(MN)2=15^2+16^2

(MN)^2=225+256

MN= $\sqrt{481}$

MN=21.93

after rounding off MN=21.9

Step-by-step explanation:

8 0

3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.