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3 years ago

Solve the following equation for the variable a.

Mathematics

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

Step-by-step explanation:

Just take b in another side by neg-

and divide by 2 , that's it.

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Help me with this please!!!

larisa86 [58]

Answer:15

Step-by-step explanation: since it’s 4 to3 and if you multiply 4 and 5 you get 20 so you multiply 3 and 5 to get 15. Hope it helps!

8 0

3 years ago

An insured's roof cost $4,000 when installed 5 years ago. It has been damaged by hail and must be replaced. The new roof will co

hoa [83]

Answer:

ACV=$4,500

Step-by-step explanation:

We have that the actual cash value (ACV) is defined as:

$ACV=\dfrac{R\times(E-C)}{E}$

Where:

$ACV =$ actual cash value

$R =$ replacement cost or purchase price of the item

$E =$ expected life of the item

$C =$ current life of the item

Then we have R=$6,000, C=5years, and to find the expected life of the item we can use the depreciating of the roof, then if the roof is depreciating $200 each year we just need to divide $4,000 by $200 to find the expected life of the roof:

$\dfrac{4,000}{200}=20$

Then the espected life of the roof is 20 years, with this result we have all the data, then:

$ACV=\dfrac{\$6,000\times (20-5)}{20}=\dfrac{\$6,000\times (15)}{20}=\dfrac{\$90,000}{20}=\$4,500$

Then the ACV is $4,500

5 0

3 years ago

A nutrition laboratory wants to construct a 95% confidence interval for the mean sodium content of "reduced sodium" hot dogs. A

tatuchka [14]

Answer:

n > 96

Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg

Step-by-step explanation:

Given;

Standard deviation r= 25mg

Width of confidence interval w= 10mg

Confidence interval of 95%

Margin of error E = w/2 = 10mg/2 = 5mg

Z at 95% = 1.96

Margin of error E = Z(r/√n)

n = (Z×r/E)^2

n = (1.96 × 25/5)^2

n = (9.8)^2

n = 96.04

n > 96

Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg

6 0

3 years ago

CAN SOMEONE HELP ME !!!!!!!!!

MissTica

Answer:

She will get <u>80mg</u> of dextromethorphan and <u>800mg</u> of guaifenesin. And the bottle last for <u>6 days</u> approximately.

Step-by-step explanation:

Given that the Robitussin DM contains dextromethorphan 10mg/5mL and gualfenesin 100mg/5mL. And we are also given that Mrs Smith took four doses and each dose is 2 teaspoons=2X5=10mL.

So, four doses=4X10=40mL.

So, dextromethorphan in 4 doses is = $\frac{10}{5}X40=80mg$

And Guaifenesin in 4 doses is = $\frac{100}{5}X40=800mg$

Dosage of medicine daily she has to take=40mL and the bottle contains 237 mL. Hence the number of days bottle last = $\frac{237}{40}=5.925$ ≈6 days approximately.

7 0

3 years ago

A particular battery claims to have a mean life of 400 hours with a standard deviation of 30 hours. Approximately what percent o

Svetach [21]

Using the normal distribution, it is found that 25.14% of the batteries will last more than 420 hours.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, we have that the mean and the standard deviation are given, respectively, by:

$\mu = 400, \sigma = 30$ .

The proportion of the batteries will last more than 420 hours is <u>one subtracted by the p-value of Z when X = 420</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{420 - 400}{30}$

Z = 0.67

Z = 0.67 has a p-value of 0.7486.

1 - 0.7486 = 0.2514.

0.2514 = 25.14% of the batteries will last more than 420 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

3 0

2 years ago