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3 years ago

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in an auditorium 1/6 of the seats are for fifth grader 1/3 are for fourth grader and the remaining students are second graders.i

f there are 96 students in the auditorium how many seconds graders are there show your work

1 answer:

DiKsa [7]3 years ago

8 0

1/2 of the students are for 5th or 4th graders. 1/2 of 96 is 48, so there are 48 second graders.

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simplify the expression. write the answer using scientific notation. (9x10^7)(7x10^9) 1.6x10^17 1.6x10^64 6.3x10^64 6.3x10^17

Daniel [21]

$The\ scientific\ notation:a\cdot10^n\ where\ 1\leq a \ \textless \ 10\ and\ n\in\mathbb{Z}\\----------------------------\\\\(9\cdot10^7)(7\cdot10^9)=(9\cdot7)(10^7\cdot10^9)=63\cdot10^{7+9}\\\\=63\cdot10^{16}=6.3\cdot10^{16+1}=6.3\cdot10^{17}$

5 0

4 years ago

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the weight M of an object on Mars varies directly as it’s weight E on Earth. A person who weighs 95 lb on Earth weighs 38 lb on

lorasvet [3.4K]

Answer:

A 100 lb person will weight 40 lb on Mars

Step-by-step explanation:

Given:

Weight on Mars directly varies as its weight on earth.

Let weight on Mars be = $M$

Let weight on Earth be = $E$

Its known that:

$M$ ∝ $E$

Thus

$M=kE$

where $k$ is constant of proportionality.

when $E=95\ lb$ , then $M=38\ lb$

thus, we have

$38=k(95)$

dividing both sides by 95.

$\frac{38}{95}=\frac{k(95)}{95}$

∴ $k=\frac{38}{95}$

Thus when $E= 100$ , $M$ would be calculated as

$M=\frac{38}{95}\times 100$

$M=\frac{3800}{95}$

∴ $M=40\ lb$ (Answer)

∴ A 100 lb person will weight 40 lb on Mars.

8 0

3 years ago

The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard devia

KengaRu [80]

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

$W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}$

The values of the cumulative distribution function of W, denoted by $\phi$ , can be found in the attached file. Lets call $y = \frac{z-70}{4}$ . We have

$0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)$

Thus

$\phi(y) = 1-0.1 = 0.9$

by looking at the table, we find that y = 1.28, therefore

$\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12$

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

8 0

3 years ago

It costs $7 for a ticket to the Pike

egoroff_w [7]

Answer:

let the number of snacks bought be x

f(x) = 7 +2.5(x)

=> f(x) is the amount you spend in total

3 0

3 years ago

No is divisible by 7 if n is divided by 2 3 4 5 the remainder is 1 what is the smallest value that n could b

kvv77 [185]

Answer: $301$

Step-by-step explanation:

Since, The LCM of numbers given numbers $2$ , $3$ , $4$ and $5$ is $60$ .

Thus, the number that gives 1 as reminder and is the multiple of 7 is $60 n + 1$

Where n is any positive integer,

Since, For $n = 5$ ,

The number is, $60\times 5 + 1 = 300 + 1 = 301$

Which is divisible by $7$ .

Thus, the required number is 301.

Note : For n = 1, 2 3 and 4, numbers are 61, 121, 181 and 241

But they are not the multiple of 7.

7 0

3 years ago

Read 2 more answers