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Allisa [31]
3 years ago
13

B) Re-write the statement 16 + 5 + 2 - 1 by including one pair of brackets

Mathematics
1 answer:
Furkat [3]3 years ago
3 0
B) 16 + (5 + 2) - 1

I don’t know what C is
You might be interested in
Four brothers each bought two hot dogs and a bag of chips at the football concession stand. If the bag of chips was $1.25, and t
IRISSAK [1]
Items bought in total:
4 brothers X two hot dogs and one bag of chips=
4X3=
12.

Hot dogs- x, chips- y.
4y + 8x= $18.20

Chips= $1.25 each so
4($1.25) + 8x= $18.20
5 + 8x= $18.20
8x= $18.20-5
8x= $13.20
x= $13.20/8 [13.20 divided by 8]
x= $1.65

This means that one hot dog= $1.65.

If you’re unsure what to write as the answer I’d suggest that you put all the working out down in your copy/notes.
3 0
2 years ago
Is 25.789 bigger then 25.879
dimulka [17.4K]
It is not bigger. 25.879 is bigger by .100
6 0
3 years ago
Which of the following function types exhibit the end behavior (PICTURE INCLUDED)
Nostrana [21]

Answer:

y=x^n; n is even and greater than zero.

The absolute value function y=|x|

Step-by-step explanation:

The power function;

y=x^n; n is even and greater than zero, is symmetric with respect to the positive y-axis.

As x\rightarrow \infty,f(x)\rightarrow \infty and as x\rightarrow -\infty,f(x)\rightarrow \infty

The absolute value function y=|x| also has the same end behavior.

The remaining functions in the options provided do not exhibit this end behavior.

See graph in attachment.

4 0
3 years ago
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
3 years ago
Enter the equivalent distance in km in the box.
devlian [24]
100 cm = 1m

To find 60 000cm, multiply by 600 :
60 000 cm = 600m                

To convert from m to km, divide by 1000 :
600m ÷ 1000 = 0.6km                        
3 0
3 years ago
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