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OleMash [197]
3 years ago
10

Alana spent $18.24 on 4.8 pounds of plant food.

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

Alana spent $18.24 on 4.8 pounds of plant food.

Which expression shows a good estimation for finding the cost of each pound of

plant food?

$

✔ 20 ÷ 5

How much did each pound of plant food cost?

$

✔ 3.80

boyakko [2]3 years ago
4 0
Dhenejjejiiiiiijnenenebenenneneejej
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Solve. x – 6 = 14 – 8 0 12 20 –12
lana [24]
The answer is 12 because first we have to simplify 14-8 to 6 so it will be
x-6=6
Then add 6 to both sides lastly simplify 6+6 to 12
So x=12
7 0
3 years ago
Write this expression in radical form t^-3/4
ZanzabumX [31]
The first thing you should know are properties of exponents to solve the problem.
 For this case the radical form is given by the writing of the expression in the form of root.
 We have then:
 t^-3/4 =4^root(t^-3)=4^root ((1)/(t^3))
answer  t^-3/4=4^root((1)/(t^3))
5 0
3 years ago
The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu
AURORKA [14]

Answer:

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21        x2`= 27

n1=  10       n2= 14

s1= 5.6       s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27)  +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 <  ( 6 )  +2.11*2.111

6- 4.4521 < u1-u2 <  ( 6 )  +5.294

- 1.5479 < u1-u2 <  10.4521

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`)  /√s1²/n1 + s2²/n2

t= (21-27)  /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0
3 years ago
50 POINTS!
padilas [110]
To double the principle the formula is
2p=p e^rt
2=e^rt
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5 0
3 years ago
What is the expression in radical form?<br><br> (4x3y2)310
sertanlavr [38]

Given:

Consider the given expression is

(4x^3y^2)^{\frac{3}{10}}

To find:

The radical form of given expression.

Solution:

We have,

(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

Therefore, the required radical form is \sqrt[5]{8y^3}\sqrt[10]{x^9}.

8 0
3 years ago
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