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2 years ago

6

will the inequality "7 multiplied by a number is greater than or equal to 21" have an open boundary point or a closed boundary p

oint on the number line?

1 answer:

ehidna [41]2 years ago

5 0

Answer:

Closed boundary point

Step-by-step explanation:

It says greater than OR equal too and having a closed boundary will indicate that 21 is apart of the inequality also.

I hope this helps

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5p coins weigh approximately 3 grams.

Elza [17]

Answer: 0.3 kg

Step-by-step explanation:

£5 = 500p

500p ÷ 5 = 100

100 x 3 = 300 grams

300 ÷ 1000 = 0.3 kg

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Simplify. (x^3 +8 ) / (x+2) <br>show work

Zielflug [23.3K]

Answer:

3x2|9x+18|+8|9x+18|3x2|9x+18|+8|9x+18|

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2 years ago

Lydia graphed ΔXYZ at the coordinates X (0, −4), Y (2, −3), and Z (2, −6). She thinks ΔXYZ is a right triangle. Is Lydia's asser

katrin [286]

Answer:

Option (4)

Step-by-step explanation:

Vertices of the triangle are X(0, -4), Y(2, -3) and Z(2, -6).

Slope of XY ( $m_{1}$ ) = $\frac{y-y'}{x-x'}$

= $\frac{-3+4}{2-0}$

( $m_{1}$ ) = $\frac{1}{2}$

Similarly, slope of XZ ( $m_{2}$ ) = $\frac{-6+4}{2-0}$

( $m_{2}$ ) = (-1)

$m_{1}\times m_{2}=-\frac{1}{2}$

Which should be (-1) if XY and XZ are perpendicular to each other.

Now we can say that XY and XZ are not perpendicular.

Therefore, Lydia's assertion that ΔXYZ is a right triangle is not correct.

Option (4) will be the answer.

5 0

2 years ago

Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow

prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

$CI=\hat{Y}\pm t_{Critical}\times S_{e}$

Where,

$\hat{Y}=$ predicted value of lead content when traffic flow is 15.

$\to df=n-1=8-1=7$

$95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\$

$=\frac{1059.8}{2}\\\\=529.9$

Calculating thet-critical value $t_{ \{\frac{\alpha}{2},\ df \}}=-2.365$

The lower predicted value $=529.9-2.365(Se)$

$463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076$

When $99\%$ of CI use as the expected lead content: $\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)$

8 0

2 years ago

Which is the iterative rule for the sequence 3, 8, 13, 18, 23, . . . . ?

Damm [24]

I'm sure this is the answer

D) an = 2 + 5n

ùwú~☆

7 0

3 years ago