The average temperature for freezing foods is less than -17°C. Which of the following best describes this situation?

Find the first, fourth, and eighth terms of the sequence. A(n)=-2 × 2^N-1

an=(−2x)(2−1)Step 1: Divide both sides by n.ann=−xna=−xnAnswer:a=−xn

7 0

3 years ago

Ratio of centimeters over inches

kolbaska11 [484]

The ratio is 0.394 cm. / 1 in.

6 0

3 years ago

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A professional organization sells term life insurance and major medical insurance. Of those who have just life insurance, 60% wi

kherson [118]

Answer:

The answer is "0.765 and 0.2353".

Step-by-step explanation:

Please find the complete question in the attached file.

In point a:

P(a substantive term only) $=0.75-0.2 =0.55$

P(major health insurance only) $= 0.45-0.20=0.25$

P(both) $= 0.20$

P(renewal) =P(insurance and renewal term only)+P (substantial and renewable health insurance only)+P (both and renew)

$=0.55\times 0.7+0.25 \times 0.8+0.2 \times 0.9 \\\\ =0.765$

In point b:

In reality, the probability of having both life and major medical insurance provided the policyholder would renew next year

$= \frac{\text{P(both and renew)}}{\text{P(renew)}}$

$=\frac{0.2\times 0.9}{0.765}\\\\=0.2353$

4 0

3 years ago

Divide. Round your answer to the nearest thousandth. 2,891.3030 - 0.728=

ira [324]

Answer:

3971.570

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

2 years ago