Janet is designing a frame for a client. She wants to prove to her client that m∠AGE ≅ m∠CHE in her sketch. What is the missing

a(t) = 210 - 5t

(1) What does t represent?

Since t is given in minutes, it stands for time.

Interpreting t as the balloon’s altitude is incorrect since a(t) represents altitude.

-------------------------------------------------------------------------------------------------

Time elapsed since the balloon starts descending

This is the correct interpretation.

---------------------------------------------------------------------------------------------------

Time for the balloon to reach ground level is incorrect because that would be a fixed moment, not a variable t.

---------------------------------------------------------------------------------------------------

(2) What does a(t) represent?

The balloon’s altitude

This is the correct answer.

-------------------------------------------------------------------------------------------------

Time since descent began is incorrect because that is represented by t.

---------------------------------------------------------------------------------------------------

Time to reach the ground is inaccurate since a(t) is altitude, not time.

---------------------------------------------------------------------------------------------------

(3) What does a(5.5) signify?

It does not represent the time to reach 5.5 feet altitude.

---------------------------------------------------------------------------------------------------

The altitude of the balloon after 5.5 minutes

This is the correct interpretation since a(5.5) is altitude when t = 5.5 minutes.

---------------------------------------------------------------------------------------------------

It is not the time needed to reach the ground.

---------------------------------------------------------------------------------------------------

5.5 minutes alone is not correct because a(t) denotes altitude, not time.

a(5.5) refers to altitude at t = 5.5 minutes.

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

Answer:

-9 idjjeninusnuwniwni

Answer:

ok __?__ equals 11 and ___ equals 3.55

Step-by-step explanation:

Start by subtracting 140 and 25 from 300, then divide by 11.95 which results in 11; subsequently, multiply 11 by 11.95 and deduct that from 135.

The test statistic (Z) is 2.5767, and the p-value of the test is 0.009975. The null hypothesis suggests that the smoking rate among students has not changed, while the alternative indicates otherwise. The z-statistic for the sampled proportion is computed, yielding z ≈ 2.5767. As we investigate whether the smoking percentage has shifted over the preceding five years, the two-tailed p-value is found to be 0.009975. This result is significant at a 99% confidence level, demonstrating substantial evidence that the percentage of smoking students has changed.