Quiz 3

Answer:

A biconditional statement integrates a conditional statement alongside its converse in an if and only if format. Two line segments are congruent if and only if their lengths are identical... A biconditional is deemed true only when both conditional statements are true. there??

Step-by-step explanation:

The ratio of women to men is 34 to 382. To find the total sales, we compute 382 multiplied by 2.50, resulting in $955. For women, the calculation is 34 multiplied by 1.25, which equals $42.50. Therefore, summing these amounts gives us $955 + $42.50 = $997.50.

Bart needs to;

Calculate 6(400), 6(10), and 6(2) ⇒ 2nd

Calculate each addend multiplied by 6 ⇒ 4th

Determine the total of (2400 + 60 + 12) ⇒ 5th

Step-by-step breakdown:

We'll review the distributive property

• a(b + c) = ab + ac
• a(b + c + d) = ab + ac + ad

Bart applied place value along with the distributive property to express the

calculation 6(412)

His calculation can be represented as 6(412) = 6(400 + 10 + 2)

We must find out what Bart needs to do to compute the product

∵ 6(412) = 6(400 + 10 + 2)

- Multiply every component inside the brackets by 6 on the right side

∴ 6(400 + 10 + 2) = (6)(400) + (6)(10) + 6(2)

∴ 6(400 + 10 + 2) = 2400 + 60 + 12

∴ He should multiply each addend by 6, then calculate the total of

  (2400 + 60 + 12)

Bart needs to;

Calculate 6(400), 6(10), and 6(2)

Multiply each addend by 6

Compute the total of (2400 + 60 + 12)

Learn more:

You can find additional information regarding place value at

Answer:

Answer and Explanation:

We have:

Population mean,

μ

=

3

,

000

hours

Population standard deviation,

σ

=

696

hours

Sample size,

n

=

36

1) The standard deviation for the sampling distribution:

σ

¯

x

=

σ

√

n

=

696

√

36

=

116

2) By the central limit theorem, the sampling distribution's expected value matches the population mean.

Thus:

The expected value of the sampling distribution equals the population mean,

μ

¯

x

=

μ

=

3

,

000

The standard deviation of the sampling distribution,

σ

¯

x

=

116

The sampling distribution of

¯

x

is roughly normal due to a sample size greater than

30

.

3) The likelihood that the average lifespan of the sample falls between

2670.56

and

2809.76

hours:

P

(

2670.56

<

x

<

2809.76

)

=

P

(

2670.56

−

3000

116

<

z

<

2809.76

−

3000

116

)

=

P

(

−

2.84

<

z

<

−

1.64

)

=

P

(

z

<

−

1.64

)

−

P

(

z

<

−

2.84

)

=

0.0482

In Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)

4) The probability of the average life in the sample exceeding

3219.24

hours:

P

(

x

>

3219.24

)

=

P

(

z

>

3219.24

−

3000

116

)

=

P

(

z

>

1.89

)

=

0.0294

In Excel: =NORMSDIST(-1.89)

5) The likelihood that the sample's average life is lower than

3180.96

hours:

P

(

x

<

3180.96

)

=

P

(

z

<

3180.96

−

3000

116

)

=

P

(

z

<

1.56

)

=

0.9406

The correct choice is C.