Tammy has ridden 54 miles of a bike course. The course is 60 miles long. What percentage of the course has Tammy ridden so far?

Three horses, call them a, b, and c, are going to race against each other. the probability that a will win the race is 1⁄5 and t

Genrish500 [490]

=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)=)

6 0

2 years ago

PLS HELP IM STUCK ON THIS ONE PROBLEM!! PLEASE SHOW STEPS

natka813 [3]

you have to solve for X is somthnking

3 0

2 years ago

What is this answer because I can't figure this out and my math teacher doesn't explain it very well

iVinArrow [24]

72+72+144 and 144 minus 360 equals 216 so half of 216 is 10. so the top and bottom are 108 degrees.

5 0

2 years ago

Read 2 more answers

I need this asap ty:p

Anastasy [175]

Answer:

(a) x = -2y

(c) 3x - 2y = 0

Step-by-step explanation:

You can tell if an equation is a direct variation equation if it can be written in the format y = kx.

Note that there is no addition and subtraction in this equation.

Let's put these equations in the form y = kx.

(a) x = -2y

y = x/-2 → y = -1/2x
This is equivalent to multiplying x by -1/2, so this is an example of direct variation.

(b) x + 2y = 12

2y = 12 - x
y = 6 - 1/2x
This is not in the form y = kx since we are adding 6 to -1/2x. Therefore, this is NOT an example of direct variation.

(c) 3x - 2y = 0

-2y = -3x
y = 3/2x
This follows the format of y = kx, so it is an example of direct variation.

(d) 5x² + y = 0

y = -5x²
This is not in the form of y = kx, so it is NOT an example of direct variation.

(e) y = 0.3x + 1.6

1.6 is being added to 0.3x, so it is NOT an example of direct variation.

(f) y - 2 = x

y = x + 2
2 is being added to x, so it is NOT an example of direct variation.

The following equations are examples of direct variation:

x = -2y
3x - 2y = 0

7 0

2 years ago

Read 2 more answers

A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi

Luden [163]

Answer:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

Step-by-step explanation:

Data given and notation

$\bar X=3.4$ represent the sample mean

$\sigma=0.4$ represent the population deviation for the sample

$n=50$ sample size

$\mu_o =3.3$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

6 0

2 years ago