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wariber [46]
3 years ago
8

( PLEASE ANSWER WILL GIVE BRAINIEST AND A THANK YOU ON PROFILE!!) On the previous problem, you found a unit rate of ounces per b

ox. Explain how you find a unit rate when given a rate.
Mathematics
2 answers:
kari74 [83]3 years ago
8 0
<span>First, I would write the rate as a fraction. Then I would divide the numerator by the denominator.</span>
Nostrana [21]3 years ago
4 0
To find<span> the </span>unit rate<span>, divide the numerator and denominator of the </span>given rate<span> by the denominator of the </span>given rate<span>. So in this case, divide the numerator and denominator of 70/5 by 5, to </span>get<span> 14/1, or 14 students per class, which is the </span>unit rate<span>.</span>
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PLZ HELP ME NEED THIS FAST WILL GIVE BRAINLIEST
OleMash [197]

Answer: l = 60 ft and w = 30 ft

P = 2(l + w)

180 = 2(l + w)

Let x = width of his yard

Let 2x = length of his yard

180=2(2x+x)

\frac{180}{2}=\frac{2(2x+x)}{2}

90=3x

\frac{90}{3} =\frac{3x}{3}

x=30ft

l = 2x = 30(2) = 60 ft

Check:

180 = 2(l + w)

180 = 2(60 + 30)

180 = 2(90)

180 = 180

LS = RS

4 0
3 years ago
The answers to A,B,C,D
scoundrel [369]

Answer:

?

Step-by-step explanation:

7 0
3 years ago
Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .
Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228

b)  "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678

e) "=T.INV(1-0.01,25)"

And we got t_{\alpha}= 2.485

f) "=T.INV(0.025,5)"

And we got t_{\alpha}= -2.571

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have \alpha/2=0.025.

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have \alpha/2=0.025.

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086

c. Central area =.99, df = 20

 For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have \alpha/2=0.005.

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845

d. Central area =.99, df = 50

  For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have \alpha/2=0.005.

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means \alpha =0.01

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got t_{\alpha}= 2.485

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means \alpha =0.025

We can use the following excel code:

"=T.INV(0.025,5)"

And we got t_{\alpha}= -2.571

8 0
3 years ago
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
I'll mark you brainlist if you answer quickly!!<br> Graph the equation 2x + 3y + z = 6
topjm [15]

Answer:

dhbiudvbe7tvderyubgydctvd7rfvd7c2ryegcfh73tr9fr4

Step-by-step explanation:

r4hfuctrfuithfy7tufhtygftfvbtyfbt53uhfbt357gfbt7fgbt7f

6 0
3 years ago
Read 2 more answers
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