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2 years ago

Please help me!!! some easy algebra

Mathematics

2 answers:

Vadim26 [7]2 years ago

6 0

I think this answer is correct We have read this question like this only

kow [346]2 years ago

4 0

Answer:

5xy-x^2t+2x7+3x^2t= 7xy+2x^2t

Step-by-step explanation:

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What is the formula to find the area of parallelogram

Fudgin [204]

A=bh will be ur answer

3 0

2 years ago

Read 2 more answers

List 3 values that would make this inequality true y + 7 > 18

blsea [12.9K]

Answer:It

It can be 12, 13, 14 and more up

Step-by-step explanation:

Is that because 12+7= 19 which is bigger than 18

3 0

2 years ago

The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school district

Bas_tet [7]

Answer:

A. Because the sample size is very large.

Step-by-step explanation:

The sample is quite large, so the t test is safe. With a sample size of at least 100, this test is going to be safe.

So the correct answer is:

A. Because the sample size is very large.

4 0

3 years ago

Rachel loves to bake cookies, but she has an old oven that has trouble maintaining a constant temperature. Assume the acceptable

denis-greek [22]

Answer:

3.00

Step-by-step explanation:

USL = 350+18 = 368

z = (USL-Mean)/6*σ = (368-350)/6*σ

σ = 3/0.9999966 = 3.00 (approx)

7 0

2 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

3 years ago