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3 years ago

8398 divide by 10 then by 100

Mathematics

1 answer:

BARSIC [14]3 years ago

8 0

Answer:

8.398

Step-by-step explanation:

If you are uncertain, you can do this with your calculator. However both numbers doing the dividing are powers of 10 so you can just move the decimal

8398 / 10 = 839.8 Move the decimal 1 place.

839.8/100 = 8.398 Move the decimal 2 places.

In both instances, move the decimal left.

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Write the equation of the line that is perpendicular to the graph of 2x+y =5, and whose y-intercept is 4.

Stells [14]

Answer and Step-by-step explanation:(MAKE BRAINLIEST PLEASE!!)

Let the quantity of understudies going to the game be s and the quantity of non-understudies going to the game be n. The all out number of individuals going to the game is 500. Address the present circumstance in a condition as s + n = 500.

Every understudy pays $5 for confirmation. Along these lines, s understudies pay 5s dollars in confirmations. Each non-understudy pays $10 for affirmation. Along these lines, n non-understudies pay 10n dollars in affirmations. Thus, the all out affirmations gathered is 5s + 10n.

The aggregate sum gathered was $4,000. So the subsequent condition is 5s + 10n = 4,000.

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3 years ago

John wants to make two frames for his artwork. One frame requires 3 1/2 feet of molding the other requires 4 2/3 feet of the sam

Damm [24]

no because if you make each fraction equal , the first would be 3 3/6 feet of molding and the other would be 4 4/6 feet and if you add those together you would get 8 1/6 which is less than 8 3/5

4 0

3 years ago

The mean age when smokers first start is 13 years old with a population standard deviation of 2 years. A researcher thinks that

lakkis [162]

Answer:

Null hypothesis: $\mu = 13$

Alternative hypothesis: $\mu \neq 13$

$z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191$

a. Reject the null hypothesis

c. There is enough evidence to support the claim

Step-by-step explanation:

Data given and notation

$\bar X=12/2$ represent the sample mean

$s=20000$ represent the standard deviation for the sample

$\sigma =2$ represent the population standard deviation

$n=30$ sample size

$\mu_o =13$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the smoking age has significantly changed since the invention of ENDS—electronic nicotine delivery systems, the system of hypothesis would be:

Null hypothesis: $\mu = 13$

Alternative hypothesis: $\mu \neq 13$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{12.2-13}{\frac{2}{\sqrt{30}}}=-2.191$

What do you conclude? Using the p-value approach

Since is a two tailed test the p value would be:

$p_v =2*P(Z$

Since the $p_v$ then the correct decision is:

a. Reject the null hypothesis

And the best conclusion is :

c. There is enough evidence to support the claim

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3 years ago

Todd is putting new oak baseboard trim in his living room. He has one piece that is 12 1/2 ft long and another that is 8 2/3 ft

Arlecino [84]

Answer:

21 1/6 ft

Step-by-step explanation:

Add the lengths together

12 1/2 + 8 2/3

Get a common denominator of 6

12 1/2 *3/3 + 8 2/3 *2/2

12 3/6 + 8 4/6

20 7/6

20 +6/6+1/6

21 1/6

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2 years ago

Write an expression that is equivalent to 3/4(5z+16).

Sphinxa [80]

Answer: 3.75z+12
For an equivalent expression, lets simplify the equation.

3/4(5z+16)
multiply both '5z' and '16' by 3/4
3.75z+12

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3 years ago

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