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yawa3891 [41]
3 years ago
11

10points! will make brainliest for correct answer -8 1/3 + (- 1/6)

Mathematics
1 answer:
pentagon [3]3 years ago
8 0

Answer: -8 1/2 I think

Step-by-step explanation:

First you make -8 1/3 a improper fraction

than you find common denominators

and you keep the sign of the biggest number

simplify if possible

You might be interested in
GETS BRAINLIEST AND 11 PTS!
vladimir2022 [97]
The answer is 70m x 10m!
3 0
3 years ago
Sofia has 25 coins in nickels and dimes in her pocket for a total of $1.65 how many of each type of coin does she have?​
Simora [160]

Answer:

She has 23 nickels and 5 dimes.

Step-by-step explanation:

Let n = the number of nickles and d = the number of dimes.

The number of each = 28. Set up the equation:

n + d = 28

Solve for n:

n = 28 - d

The total amount is $1.65 or 165 cents:

5n + 10d = 165

Substitute:

5(28 - d) + 10d = 165

140 - 5d + 10d = 165

140 + 5d = 165

5d = 165 - 140

5d = 25

d = 5, the number of dimes.

Solve for n:

n = 28 - d

n = 28 - 5 = 23, the number of nickles.

Proof:

5n + 10d = 165

5(23) + 10(5) = 165

115 + 50 = 165

165 = 165

Hope this Helps! Have an Awesome Day!!! <3

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Can someone help me with these?​
konstantin123 [22]

Answer:

Step-by-step explanation:

4 0
3 years ago
Can someone please explain how to solve this step by step with answer? Thank you.
lidiya [134]

a)

well, she put 4000, and she earned in interest 960, so her accumulated amount is just their sum, 4960.

b)

now, it doesn't say, so we're assuming is <u>simple interest</u>, as opposed to compound interest.


\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill&960\\ P=\textit{original amount deposited}\dotfill & \$4000\\ r=rate\to r\%\to \frac{r}{100}\dotfill\\ t=years\dotfill &3 \end{cases} \\\\\\ 960=(4000)(r)(3)\implies \cfrac{960}{(4000)(3)}=r\implies 0.08=r \\\\\\ \stackrel{\textit{converting to percentage}}{r=0.08\cdot 100}\implies r=\stackrel{\%}{8}


c)

let's make the rate 1% greater then, and check


\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill\\ P=\textit{original amount deposited}\dotfill & \$4000\\ r=rate\to \stackrel{8+1}{9\%}\to \frac{9}{100}\dotfill&0.09\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(4000)(0.09)(3)\implies I=1080 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{at 9\%}}{1080}-\stackrel{\textit{at 8\%}}{960}\implies \stackrel{\textit{that much more}}{120}

4 0
3 years ago
A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio
Reptile [31]

Answer:

The probability is P(|p-\^{p}| >  0.03)  =   0.0040

Step-by-step explanation:

From the question we are told that

   The population proportion is p =  0.07

   The mean of the sampling distribution is   \mu_p =  0.07

   The sample size is n = 600

Generally the standard deviation is mathematically represented as

     \sigma_p  =  \sqrt{\frac{p (1 -p)}{n} }

=>    \sigma_p  =  \sqrt{\frac{0.07(1 -0.07)}{600} }    

=>    \sigma_p  =  0.010416    

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

      P(|p-\^{p}| >  0.03) =  1 - P(|p -\^{p}| \le 0.03)

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(-0.03 \le p -\^{p} \le 0.03 )

Now  add p  to  both side of the inequality

=>   P(|p-\^{p}| >  0.03)  =  1 -  P( 0.07-0.03  \le \^{p} \le 0.03+ 0.07 )

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(0.04 \le \^{p} \le 0.10 )

Now  converting the probabilities to their respective standardized score

=> P(|p-\^{p}| >  0.03)  =  1 -  P(\frac{0.04 - 0.07}{0.010416}  \le Z \le \frac{0.10 -0.07}{0.010416}  )

=> P(|p-\^{p}| >  0.03)  =  1 -  P(-2.88  \le Z \le 2.88 )

=>   P(|p-\^{p}| >  0.03)  =   1 - [P(Z \le 2.88) - P(Z \le -2.88)]

From the z-table  

       P(Z \le 2.88)  =  0.9980

and

       P(Z \le -2.88)  = 0.0020

So

     P(|p-\^{p}| >  0.03)  =   1 - [0.9980 - 0.0020]

=>   P(|p-\^{p}| >  0.03)  =   0.0040

     

7 0
3 years ago
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