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oksian1 [2.3K]
3 years ago
13

I don’t understand

Mathematics
1 answer:
Alchen [17]3 years ago
6 0
It’s in scientific, currently. So you want to put it back the way you’d write it normally. Move the decimal two places to the right (bc x 10^2) to get 338.55
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Summer is having a pool party at her house In the ice chest me are of lemonade and 13 bottles of water If she randomly grabs abo
zlopas [31]
Probability  = 

       (number of ways it can come out the way you want it to)
divided by
       (total number of ways it can come out).

Total number of ways it can come out =

                    number of bottles in the cooler

                 = (10 + 15 + 13) = 38 .

Number of ways it can come out the way you want it to =

                   (soda + lemonade)

                = (10 + 15) = 25 .

Probability of coming out the way you want it to =

                     25 / 38  =  about 65.8 % .

7 0
2 years ago
Solve 9/10=z/8 . Round to the nearest tenth.
kompoz [17]
\boxed {Question}
\frac{9}{10} =  \frac{z}{8}

\boxed {Cross \  multiply}
10z = 8 \times 9

\boxed {Evaluate}
10z = 72
z = 72 \div 10
z = 7.2

\Longrightarrow \bf \ Answer \ : \ z = 7.2

4 0
3 years ago
Simplify for a lot of points<br> (1-sin^2x)(1-tan^2x)
notka56 [123]
1/sin^2x-1/tan^2x=
1/sin^2x-1/ (sin^2x/cos^2x)<<sin tan= sin/cos>>
= 1/sin^2x- cos^2x / sin^2x
= (1- cos^2x) / sin^2x <<combining into a single fraction>>
sin^2 x / sin^2x <<since 1- cos^2 x sin^2 x

=1
this simplifies to 1.
6 0
3 years ago
1/4 ,1/2, 1, 2, solve for a10
LUCKY_DIMON [66]

Answer: {a}_{10} = 128

Step-by-step explanation:

3 0
3 years ago
For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

3 0
3 years ago
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