Pie Chart 3: Drinking Water Services (of 7.8 Billion People)
1 answer:
Answer:
access to safe drinking water is a fundamental need and human right. Securing access for all would go a long way in reducing illness and death, especially among children. “Safely managed” drinking water services represent an ambitious new rung on the ladder used to track progress on drinking water. Since 2000, 2 billion people have gained access to safely managed services (i.e., accessible on-premises, available when needed, and free from contamination). In 2020, 5.8 billion people used safely managed services and a further 2 billion people used basic services ( improved sources within 30 minutes per round trip to collect water).
However, 771 million people still lacked even a basic level of service, including 282 who used a “limited” water service (improved source from which water collection exceeds 30 minutes), 367 million who used unimproved sources and 122 million who still collected drinking water directly from rivers, lakes, and other surface water sources. The data reveal pronounced disparities, with the poorest and those living in rural areas least likely to use a basic service. In most countries, the burden of water collection continues to fall mainly to women and girls.
Step-by-step explanation:
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Answer:
(a) Decreasing on (0, 1) and increasing on (1, ∞)
(b) Local minimum at (1, 0)
(c) No inflection point; concave up on (0, ∞)
Step-by-step explanation:
ƒ(x) = x² - x – lnx
(a) Intervals in which ƒ(x) is increasing and decreasing.
Step 1. Find the zeros of the first derivative of the function
ƒ'(x) = 2x – 1 - 1/x = 0
2x² - x -1 = 0
( x - 1) (2x + 1) = 0
x = 1 or x = -½
We reject the negative root, because the argument of lnx cannot be negative.
There is one zero at (1, 0). This is your critical point.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.
(1) x = ½
ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1
ƒ'(x) < 0 so the function is decreasing on (0, 1).
(2) x = 2
ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂
ƒ'(x) > 0 so the function is increasing on (1, ∞).
(b) Local extremum
ƒ(x) is decreasing when x < 1 and increasing when x >1.
Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.
(c) Inflection point
(1) Set the second derivative equal to zero
ƒ''(x) = 2 + 2/x² = 0
x² + 2 = 0
x² = -2
There is no inflection point.
(2). Concavity
Apply the second derivative test on either side of the extremum.
The function is concave up on (0, ∞).
