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Brut [27]
3 years ago
11

How many solutions exist for the given equation?

Mathematics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer: Infinitely many solutions when we graph this it comes out as one straight line also try using Desmos it helps with equations like this by graphing them for you! -Your friend, Bill Cipher

Step-by-step explanation: Have a great Valentines day <3

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Jim ordered a sports drink and three slices of pizza for $8.50. His friend ordered two sports drinks and two slices of pizza for
kramer

Answer:

total is 16.50

Step-by-step explanation:

add 8.50+8.00

7 0
3 years ago
Find the width of the rectangular prism when the surface area is 208 square centimeters.
ozzi
SA=2(lw+wh+lh) This is the formula for finding the surface area of a rectangular prism, where SA is surface area, l is length, w is width, and h is height.

208=2(lw+wh+lh)
104=lw+wh+lh Here, I divided both sides by 2 to get ride of the 2.

Now, I used prime factorization to find out all the prime factors of 104, which are 2, 2, 2, and 13. Since rectangular prisms only have 3 dimensions, I needed to combine two of the prime factors. In this case, I can either combine 2 of the 2s to get 2, 4, and 13 or I can combine 13 with one of the 2s to get 26, 2, and 2.

If my dimensions were 2, 4, and 13...
my surface area would be 172 sq cm.

If my dimensions were 2, 2, and 26...
my surface area would be 208 sq cm.

Hence, the width of the rectangular prism when the surface area is 208 square centimeters can be either 2 or 26.
7 0
3 years ago
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Tcecarenko [31]

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408

#Define Y:

Y~(11.04,5.7408)

P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291

Hence, the probability of 12 or fewer is 0.8291

b. The  probability that 5 or more fish were caught.

#Using normal approximation:

P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above P(X\geq 5)=0.9968 and a ,P(X\leq 12)=0.7291

P(5\leq X\leq 120\approx P(4.5\leq Y\leq  12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259

Hence, the probability of between 5 and 12 is 0.7259

4 0
3 years ago
The expression 937(1+x) gives the markup price of a computer, where x is the percent of the markup written in decimal form. Whic
jarptica [38.1K]
The expression is 203.422+ X
6 0
3 years ago
Can someone help me please ?
sweet-ann [11.9K]
I hope this helps you

4 0
3 years ago
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