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bulgar [2K]
4 years ago
7

Which of the figures appear to be congruent?

Mathematics
1 answer:
krek1111 [17]4 years ago
3 0
A picture? Do you have a picture of the work
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Liz wants to find the quotient of 63 and 0.15. what is the first step she should take?
Veseljchak [2.6K]

Question

Liz wants to find the quotient of 63 and 0.15. what is the first step she should take?

Answer:

→(Well First 63 ÷ 0.15 = 420) ←

A: divide 63 and 0.15 by 10 Would be my Answer

To add two decimal numbers, first check if they have the same number of digits to the right of the decimal point. If they don't, add zeros to the right of one of the numbers until they do. Then, write one number on top of the other, lining up the decimal points vertically.

xXxAnimexXx

5 0
3 years ago
What is run pattern of j, f, m, a, m
Dafna1 [17]
Those are some of the first letters of the month. j = January, f = Feb, m = March, a = April, and m = May.

Happy studying ^-^
5 0
3 years ago
The new figure after a transformation is performed is called the __________.
Luda [366]

we know that

A figure before the transformation is called pre-image and the figure after a transformation is called image

therefore

<u>the answer is the option D</u>

Image

3 0
4 years ago
Read 2 more answers
How to show these 2 problems are inverses
nataly862011 [7]
\bf \begin{cases}&#10;f(x)=\sqrt[3]{7x-2}\\\\&#10;g(x)=\cfrac{x^3+2}{7}&#10;\end{cases}\\\\&#10;-----------------------------\\\\&#10;now&#10;\\\\&#10;f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2}&#10;\\\\\\&#10;\sqrt[3]{x^3}\implies x\\\\&#10;-----------------------------\\\\&#10;or&#10;\\\\&#10;g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7}&#10;\\\\\\&#10;\cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x

thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other
8 0
3 years ago
The General Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours
DerKrebs [107]

Answer:

Yes, we can conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

Step-by-step explanation:

H0 : σ² = 2.66²

H1 : σ² < 2.66²

X²c = (n - 1)*s² ÷ σ²

sample size, n = 40

Sample standard deviation, s = 1.9

X²c = ((40 - 1) * 1.9²) ÷ 2.66²

X²c = 140.79 ÷ 7.0756

X²c = 19.897

Using a confidence level of 95%

Degree of freedom, df = n - 1 ; df = 40 - 1 = 39

The critical value using the chi distribution table is 25.6954

Comparing the test statistic with the critical value :

19.897 < 25.6954

Test statistic < Critical value ; Reject the Null

Hence, we can conclude that the population standard deviation of TV watching times for teenagers is less than 2.66

8 0
3 years ago
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