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melamori03 [73]
3 years ago
9

The time in Rio is three hours behind London.

Mathematics
2 answers:
Debora [2.8K]3 years ago
6 0

Answer:

1:00 am

Step-by-step explanation: if it is 11:00 add the 5 hours because NYC is 5 hours behind London so when you add them you get to 4:00 am then since Rio is three hours behind you subtract 3 hours from 4:00 am and you get 1:00 am.

Anvisha [2.4K]3 years ago
3 0
The time is 1am , add 2hours to 11
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Two numbers have a distance of 6 units from 0 on a number line. The numbers can be graphed on the number line as points A and B.
lara [203]

Answer:

6

Step-by-step explanation:

7 0
3 years ago
Find the average rate of change of each function on the interval specified.
ExtremeBDS [4]

The average rate of change (AROC) of a function f(x) on an interval [a, b] is equal to the slope of the secant line to the graph of f(x) that passes through (a, f(a)) and (b, f(b)), a.k.a. the difference quotient given by

f_{\mathrm{AROC}[a,b]} = \dfrac{f(b)-f(a)}{b-a}

So for f(x) = x² on [1, 5], the AROC of f is

f_{\mathrm{AROC}[1,5]} = \dfrac{5^2-1^2}{5-1} = \dfrac{24}4 = \boxed{6}

4 0
3 years ago
a parking lot has total of 60 cars and trucks. the ratio of cars to trucks is 7:3 how many cars are in the parking lot. How many
Lerok [7]
7+3=10
60/10=6
6x7=42
6x3=18
there is 18 trucks in the parking lot
7 0
3 years ago
Graph the Quadratic Function given.<br> y = -x^2 - 4x - 3
Dovator [93]

Answer:

Concave down

Y-intercept: (0,-3)

X-intercepts: (-1,0) (-3,0)

Vertex: (-2,1)

6 0
3 years ago
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
3 years ago
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