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Ipatiy [6.2K]
2 years ago
5

Please help now i have a test

Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
8 0

Answer:

-1.75 or -1 3/4

Step-by-step explanation:

each line is 1/4 since you need 4 lines to get to the next number.

the number falls on the third line out of 4 lines

vova2212 [387]2 years ago
3 0
1-7852 -446477204oirp
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EMILY COLLECTED $950 SELLING GIRL SCOUT COOKIES ALL DAY SATURDAY. EMILY'S TROOP COLLECTED 10 TIMES AS MUCH AS SHE DID. HOW MUCH
Katyanochek1 [597]
9,500  just add the 0 always add 0 if you multiply by 10. 00 for 100 and 000 for 1000
7 0
3 years ago
Read 2 more answers
Simplify<br> (3/2x+5) + (5/x-5)
tia_tia [17]

Answer:

(3x + 10) / (2x + 5)(x - 5).

Step-by-step explanation:

(3/2x+5) + (5/x-5)

= [3(x - 5) + 5(2x + 5) ] /  [ (2x + 5)(x - 5)]

= 3x - 15 + 10x + 25 /  (2x + 5)(x - 5)

= 13x + 10 / (2x + 5)(x - 5).

6 0
3 years ago
How much is one melon if 3 sell for $1.98 ?
Sonbull [250]

Answer: $0.66


Step-by-step explanation: Just divide 3 / $1.98 and you will get $0.66.

Fairly simple, But the only thing I don't understand is why a melon is on sale for $0.66. Thats CHEAP! Give me 100 :)


8 0
2 years ago
Read 2 more answers
in 2018 GRE scores were normally distributed with a mean of 303, and standard deviation of 13. Standford University Graduate sch
eduard

Answer:

The minimum score needed to be considered for admission to Stanfords graduate school is 328.48.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 303, \sigma = 13

What is the minimum score needed to be considered for admission to Stanfords graduate school?

Top 2.5%.

So X when Z has a pvalue of 1-0.025 = 0.975. So X when Z = 1.96

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

1.96 = \frac{X - 303}{13}

X - 303 = 13*1.96

X = 328.48

The minimum score needed to be considered for admission to Stanfords graduate school is 328.48.

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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