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3 years ago

How How many solutions does the nonlinear system of equations graphed below have?

1 answer:

4 0

Upon looking, you should see 4 points where the graphs intersect, so there are 4 solutions to the non-linear system of equations.

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Ram has three times as many Rupees to coins as he has rupees 5 coins if he has in all a sum of rupees 77 how many coins of Rupee

BabaBlast [244]

9514 1404 393

Answer:

21 coins of ₹2

Step-by-step explanation:

Let x represent the number of ₹2 coins, and y the number of ₹5 coins. Then the total value of the coins is ...

2x +5y = 77

and the relationship between numbers of coins is ...

x = 3y

Substituting for x, we have ...

2(3y) +5y = 77

y = 77/11 = 7 . . . . simplify, divide by the coefficient of y

x = 3(7) = 21 . . . . find x from the second equation

Ram has 21 of the ₹2 coins.

8 0

3 years ago

Which of the following is the solution to 4|x+2|>=16. HELP ME RIGHT NOW

m_a_m_a [10]

Answer:

yes

Step-by-step explanation:

4 0

3 years ago

HELP WILL MARK BRAINLIEST

Trava [24]

Look at the attached picture⤴

Hope this will help u...

7 0

3 years ago

Read 2 more answers

Write the first six terms of the geometric sequence with the first term 6 and common ratio 1/3

Alja [10]

Answer:

6, 2, 2/3, 2/9, 2/27, 2/81

Step-by-step explanation:

The nth term of a geometric progression is expressed as;

Tn = ar^n-1

a is the first term

n is the number of terms

r is the common ratio

Given

a = 6

r = 1/3

when n = 1

T1 = 6(1/3)^1-1

T1 = 6(1/3)^0

T1 = 6

when n = 2

T2= 6(1/3)^2-1

T2= 6(1/3)^1

T2 = 2

when n = 3

T3 = 6(1/3)^3-1

T3= 6(1/3)^2

T3= 6 * 1/9

T3 = 2/3

when n = 4

T4 = 6(1/3)^4-1

T4= 6(1/3)^3

T4= 6 * 1/27

T4 = 2/9

when n = 5

T5 = 6(1/3)^5-1

T5= 6(1/3)^4

T5= 6 * 1/81

T5 = 2/27

when n = 6

T6 = 6(1/3)^6-1

T6= 6(1/3)^5

T6= 6 * 1/243

T6 = 2/81

Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81

6 0

3 years ago

A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi

34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{X}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{X}=\sqrt{n}\sigma$

It is provided that:

$\mu=\$27\\\sigma=\$18\\n=310$

As the sample size is quite large, i.e. n = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

$P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z$

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let s denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding z-value is, -2.33.

Compute the value of s as follows:

$z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57$

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0

3 years ago