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3 years ago

Write a proportion to find how many points x a student needs to score on a test worth 50 points to get a test score of 78%. $$

2 answers:

7 0

78% of 50 points = 0.78 x 50
78% of 50 points = 39

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Answer: The student needs to score 35 points.
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Brums [2.3K]3 years ago

6 0

Answer: The student needs to score 35 points.

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What's a real number between -1/10 and 0?

vova2212 [387]

Answer:

The answer is at the bottom!!

Step-by-step explanation:

And All of those Natural Numbers will be starting with 1 so we will be having other Numbers as well on which No Number will be mapped like 2 or 211 or 79 So This Means Set of Natural Numbers is Grater then Real Numbers Between 0 and 1. So Set of Real Numbers Between 0 and 1 is Countably Infinite.

Hope this helps!!

4 0

2 years ago

John and Jeremy are the local names for two lighthouses that guard a particularly dangerous part of the coast. John blinks every

Likurg_2 [28]

John blinks every 6 seconds, Jeremy every 4 seconds, so they both will blink next on a common factor of 6 and 4, or an LCD, that'd be 12 seconds later, so at 00:00:12.

8 0

3 years ago

What is the solution of the equation 4.5+m=10

drek231 [11]

Answer:

The first thing that we need to do is subtract 4.5 from both sides because that would removed the 4.5 from the left side and leave us with m by itself. So we get the following $m=5.5$ and that is our solution.

<u><em>Hope this helps! Let me know if you have any questions</em></u>

6 0

1 year ago

Read 2 more answers

A cooler has a temperature of 32 degrees Fahrenheit. A bottled drink is placed in the cooler with an initial temperature of

FinnZ [79.3K]

Answer:

$k \approx 0.44$

Step-by-step explanation:

Given function:

$f(t) = (ce)^{-kt}+32$

As per question statement:

Initial temperature of bottle is 70 $^\circ F$ .

i.e. when time = 0 minutes, f(t) = 70 $^\circ F$

$70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38$

After t = 3, f(t) = 42 $^\circ F$

$42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44$

Hence, the value is:

$k \approx 0.44$

8 0

3 years ago

a sleeping bag weighs 8 pounds. your backpack and sleeping bag together weigh 31 pounds. write an equation to determine whether

Luda [366]

31-8=23 pounds for the backpack.

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3 years ago