Help me out pleasee!!!!!!

Mathematics

2 answers:

egoroff_w [7]4 years ago

7 0

The answer would be C

larisa86 [58]4 years ago

3 0

Answer:

The answer is C. 61.92

Step-by-step explanation:

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Show your solution follow the steps<br>need an answer, please need an answer, please.

Alex777 [14]

Answer:

after calculating I got this answer.

hope this answer helps you dear...take care

8 0

3 years ago

HELP ASAP

Mumz [18]

<span>$15000 compounded continuously at a rate of 3% for 11 years will yield a final amount of $20864.52. The equation for continuously compounded interest A = Pe^rt where A=End Amount P=Inital Amount e=eulers number (approximately 2.7) r=growth rate per year and t= time. So for your problem, p= 15000 r= 3 and t= 11. Plug that into your calculator and you'll get $20864.52.</span>

5 0

3 years ago

Is 18×<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B5%7D%20" id="TexFormula1" title=" \frac{4}{5} " alt=" \frac{4}{5} "

Elden [556K]

Fraction means a part of something

we see
18 times a fraction is bigger or smaller or equal to 18

18 times a fraction means take a part ouf of 18
in this case
18 times 4/5=72/5=14.4
14.4 is less than 18
when you take a fraction of something, it becomes smaller, like if you take a slice of pizza or a fraction of the pizza, it becomes smaller

lesss

8 0

3 years ago

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le

CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁: μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

$Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }$