PLEASE HELP IN ONE MINUTE. WILL MARK BRAINLEST

If $8,000 is invested at 5 % annual interest compounded

poizon [28]

8000 (1.025)^12

$10,759

3 0

3 years ago

3-(4-2) 5 3.4-2.5 11-4-2 2015 11 (42) (7-3)-(4-2)7-3-4-2

KATRIN_1 [288]

Answer: look below :)

Step-by-step explanation:

5 < 11 < 42 < 2015

5 is less than 11 is less than 42 is less than 2015

5/

1

< 11/

1

< 42/

1

< 2015/

1

8 0

2 years ago

stacy buys 3 cds in a set for $29.98. She save $6.44 by buying the set instead of buying the individual cds. if each cd costs th

Natalka [10]

29.98+ 6.44 = 36.42 (cost with out the discount)
36.42/3 = 12.14

Answer-
12.14

4 0

4 years ago

Read 2 more answers

Jasmine is saving to buy a bicycle. The amount she has saved is shown in the table. What is the function describes the amount A,

zubka84 [21]

Answer:

A = 15t +15

Step-by-step explanation:

The amounts have a common difference of $15, so that is apparently the amount Jasmine is saving each week. Week 1, however, is $15 more than $15×1. The function ...

A = 15t +15 . . . . dollars

seems to fit the data.

8 0

3 years ago

Read 2 more answers

The desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5. To test whether the true average

Reil [10]

Answer:

a) Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

|t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Step-by-step explanation:

Step( i ):-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

Level of significance ∝ = 0.01

Step(ii):-

The test statistic

$t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }$

On calculation , we get

t = -3.33

|t| = |-3.33| = 3.33

Step(iii):-

P - value

The degrees of freedom γ = n-1 = 16-1 =15

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

Condition(i)

P - value < ∝ then reject H₀

Condition(ii)

P - value > ∝ then Accept H₀

we observe that 0.001 < P < 0.005

P- value < 0.01

we rejected H₀

(or)

The tabulated value = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

Conclusion:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

4 0

3 years ago