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Lynna [10]
2 years ago
12

The normal temperature range for Yuma, AZ on January 11 is at least 48 degrees but no higher than 64 degrees Fahrenheit. Which i

nterval describes the normal temperature in Yuma, AZ on January 11?
Mathematics
1 answer:
WITCHER [35]2 years ago
7 0

Answer: Our required interval for temperature is 48\leq x\leq 64

Step-by-step explanation:

Since we have given that

The normal temperature range for Yuma, on Janurary is atleast = 48 degrees

and the highest temperature on that day is no higher than 64 degrees.

Let the temperature in Yuma be 'x'.

So, we know that for atleast means it can 48 degrees or more, and no higher than means it should be equa to or less than 64 degrees.

Mathematically, it is expressed as

48\leq x\leq 64

Hence, our required interval for temperature is 48\leq x\leq 64

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MariettaO [177]

Answer:

1.The ratio of dogs to cats is 3 to 4

2. For every 3 dogs there are 4 cats.

Step-by-step explanation:

5 0
2 years ago
Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How
Scorpion4ik [409]

let's check how much is it after 2 years firstly.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275


Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

4 0
3 years ago
Please help and explain.
Ludmilka [50]

Answer:

It standard form is 25000

4 0
2 years ago
Read 2 more answers
What are the zeros of (x-2)(x^-9)
Nostrana [21]
I suspect you may have slipped up as you copied the question. Exactly as you wrote it, it has a single zero, at x= 2 .
8 0
3 years ago
PLEASE HELP WITH THESE 4 PROBLEMS
Marizza181 [45]

Answer:

  • y = -5/2x + 5
  • y = 2x -1
  • y = 24/11
  • y = 4x + 4

Step-by-step explanation:

To solve for y, subtract the x-term and divide by the coefficient of y.

5x +2y = 10

  2y = -5x +10

  y = -5/2x + 5

__

2y -4x = -2

  2y = 4x -2

  y = 2x -1

__

7y +4y = 24

  y = 24/11

We assume there is a typo, but cannot tell what it is. Use the same method as described and demonstrated for the other problems.

__

-2x +1/2y = 2

  1/2y = 2x +2

  y = 4x +4

8 0
3 years ago
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