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AnnZ [28]
2 years ago
10

Jim has $700 in savings at the beginning of summer. He wants to have at least

Mathematics
2 answers:
Julli [10]2 years ago
8 0

Answer:

Jim has 700 and wants 300 left. he can spend 400. 25 every weekend. divide 400 by 25 and you get 16. the answer is 16

Step-by-step explanation:

sattari [20]2 years ago
4 0

Answer:

16 weekends

Step-by-step explanation:

700-300=400

400/25=16

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It is expected that 563 quadrillion thermal units of Btu (British thermal units) of energy will be
Evgesh-ka [11]

Answer:

the expected rate of change in consumption is 11.83

Step-by-step explanation:

The computation of the expected rate of change in consumption is shown below:

= (Expected quadrillion thermal units - 2003 quadrillion thermal units) ÷ (difference in years)

= (563 - 421) ÷ (2015 - 2003)

= 142 ÷ 12

= 11.83

hence, the expected rate of change in consumption is 11.83

We simply applied the above formula so that the correct value could come

And, the same is to be considered

3 0
3 years ago
Blake said 4/5+1/3 =5/8 does his answer make sense explain. please help
Tatiana [17]
Hey there!

The answer is no because whenever you add or subtract fractions, you find the common denominator(the bottom of the fraction)

Corret way:
The common denominator is 15 (5*3)
Multiply numerator(top) and the denominator by 3 
4/5*3*3= 12/15
Multiply numerator and the denominator by 5
1/3*5/5= 5/15
So the equation is now
12/15+5/15= 17/15 or 1 2/15

Hope this helps! :)
7 0
3 years ago
Which graph represents the solution set of the system of inequalities?
Lyrx [107]
The answer to this is D
4 0
3 years ago
Read 2 more answers
Pls help ASAP!! Pleaseee now!!
kap26 [50]
4/5 x 7= 28/5= 5 3/5
6 0
3 years ago
A random sample of 121 automobiles traveling on an interstate showed an average speed of 73 mph. from past information, it is kn
tatiyna

121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range


(73 - 2 \cdot 11, 73 + 2 \cdot 11) = (51,95)


The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.


The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:


\sigma = \dfrac{ 11}{\sqrt{121}} = 1


So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range


(73 - 2 \cdot 1, 73 + 2 \cdot 1) = (71,75)


That's probably the answer they're looking for.



4 0
3 years ago
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