All categories

3 years ago

Rita had $9.00. She spent $3.67 on a magazine.

Mathematics

2 answers:

Ket [755]3 years ago

8 0

She has $5.33 left

$9.00
-$3.67
————
$5.33

jeka943 years ago

3 0

Answer:

the answer is 5.33

Step-by-step explanation:

9.00

-367

--------

5.33

You might be interested in

If a sample of 40 cars is selected, estimate the number of cars traveling faster than 70 mph.

Umnica [9.8K]

Answer:

20ish

Step-by-step explanation:

I would need more info abt the question to answer it more exactly

5 0

2 years ago

Read 2 more answers

Find the discriminant of the following equation:

vladimir1956 [14]

Option 1

The discriminant of given equation is 116

Solution:

We have to find the discriminant of given equation

Given equation is:

$f(x) = 4x^2+6x-5$

To find discriminant,

$4x^2+6x-5 = 0$

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0$

Discriminant is given by:

$D = b^2-4ac$

On comparing the given quadratic equation $4x^2+6x-5 = 0$ with general quadratic equation $ax^2+bx+c = 0$

a = 4

b = 6

c = -5

Substituting in above formula,

$D = 6^2-4(4)(-5)\\\\D = 36-16(-5)\\\\D = 36+80\\\\D = 116$

Thus discriminant of given equation is 116

8 0

3 years ago

How to solve this please help me 2/3×1/2(3/4÷1 2/5)

Damm [24]

Answer:

0.624999999(repeating) so i would round it to 0.625 or 0.63 (both work and are rounded, its really up to wherever your teacher wants you to round it to if they want you to round it)

Step-by-step explanation:

2/3 x 1/2 (3/4 ÷ 2/5)=

simplify the parentheses (3/4 ÷ 2/5) to 1.875

2/3 x 1/2 x 1.875=

simplify 2/3 x 1/2 to 0.333333(repeating)

0.3333333333 x 1.875 = 0.6249999999(repeating)

round it to 0.625 or 0.63 (which ever one your teacher wants you to round it to)

3 0

4 years ago

The answer to the question

Yuri [45]

The required equation of line parallel to given line is: $y = \frac{5}{6}x-8$

Step-by-step explanation:

Given equation of line is:

$y = \frac{5}{6}x+1$

As the equation of line is in slope-intercept form, the co-efficient of x s the slope

Let m1 be the slope of given line

m1= 5/6

Let m2 be the slope of new line

As the slopes of two parllel lines are equal, so

m1 = m2

m2 = 5/6

The slope intercept form is:

$y= m_2x+b$

Put m2 = 5/6 in equation

$y = \frac{5}{6}x+b$

Putting the point in the equation

$2 = \frac{5}{6}(12)+b\\2=10+b\\b = 2-10\\b = -8$

Putting the value of b, we get

$y = \frac{5}{6}x-8$

Hence,

The required equation of line parallel to given line is: $y = \frac{5}{6}x-8$

Keywords: Slope intercept form, equation of line

Learn more about equation of line at:

#LearnwithBrainly

4 0

3 years ago

The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6

IrinaK [193]

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$ , in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$ .

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$ .

This is P(X = 4). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0

3 years ago