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3 years ago

Which of the following numbers are rational 25/9, -25, 0.333...,

Mathematics

2 answers:

Sav [38]3 years ago

6 0

......The answer is -25.....

marshall27 [118]3 years ago

5 0

Whole numbers, integers, fractions, terminating decimals and repeating decimals are all rational numbers.

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Which expression is equivalent to the square root of -80?

Viktor [21]

Answer:

c is the correct answer

5 0

3 years ago

Help please very confused.

Nat2105 [25]

Answer:

12.

Step-by-step explanation:

Replace k in the radical by -2:

So we have √(-72*-2)

= √144

= 12.

7 0

3 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

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, where t

ratelena [41]

Answer with Step-by-step explanation:

We are given that height of projectile after t seconds is given by

$h(t)=96t-16t^2$

a.h(t)=144 ft

$144=96t-16t^2$

$16t^2-96t+144=0$

$t^2-6t+9=0$

$t^2-3t-3t+9=0$

$t(t-3)-3(t-3)=0$

$(t-3)(t-3)=0$

t-3=0

t=3

After 3 s, the height of the project will be 144 feet in the air.

b.h(t)=0

$96t-16t^2=0$

$16t(6-t)=0$

$16t=0\implies t=0$

$6-t=0\implies t=6$

At t=0, the initial position of projectile

At t=6 s , the projectile will hit the ground.

6 0

3 years ago

Suppose $1000 is invested at a rate of 13% per year compounded monthly. (Round your answers to the nearest cent.)

masya89 [10]

Answer:

a. $1010.83

b.$1066.77

c. $1138.00

d.$13,269.22

Step-by-step explanation:

Given the annual rate as 13%(compounded monthly) and the principal amount as $1000.

a. #first we calculate the effective annual rate;

$i_m=(1+i/m)^m-1\\\\i_{12}=(1+0.13/12)^{12}-1=0.1380$

The compounded amount after 1 month is therefore:

$P_1=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_1=1000(1+0.1380)^{1/12}\\\\P_1=1010.83$

Hence, the principle after one month is $1010.83

b. The principal after 6 months:

-From a above we have the effective annual rate as 0.1380 and our time is 6 months:

$P_{6m}=P(1+i_m)^n, \ n=6m, P=1000, i_m=0.1380\\\\P_{6m}=1000(1+0.1380)^{6/12}\\\\=1066.77$

Hence, the principal after 6 months is $1066.77

c.The principal after 1 year:

-From a above we have the effective annual rate as 0.1380 and our time is 12 months:

$P_{1y}=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_{1y}=1000(1+0.1380)^{12}\\\\P_{1y}=1138$

Hence, the principal after 1 year is $1138.00

d. The principal after 20years:

-From a above we have the effective annual rate as 0.1380 and our time is 20yrs:

$P_{20y}=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_{20y}=1000(1+0.1380)^{12}\\\\P_{20y}=13269.22$

Hence, the principal after 20 years is $13,269.22

3 0

3 years ago