What is k(-4) equal to?<br><br> K:m => 7+m^2<br><br> K(-4) = [1]

Y_Kistochka [10]

Circle Circumference= 2πr

C= 2(3.14)(24)
C= (6.28)(24)
C= 150.72 inches

ANSWER: If we round up, the estimated circumference is 151 inches.

Hope this helps! :)

7 0

3 years ago

Read 2 more answers

Select the two figures that are sThe tree in your friend's yard casts a shadow that is 27 feet long. Your friend is 5 feet tall

Sphinxa [80]

Answer: 30 ft tall

Step-by-step explanation: X= the height of the tree in feet

27/x = 4.5/5

(4.5)(x) = (27)(5)

(4.5)(x) = 135

divide by 4.5 to isolate x

135/4.5 = 30

x = 30

the tree is 30 ft tall

3 0

4 years ago

11. m2 = 34® Find m 24 and explain how you know,

olchik [2.2K]

Answer:

sarang heyo arigatso

Step-by-step explanation:

Sorikaniya

7 0

3 years ago

In a field study, a scientist was predicting the insect population. The initial amount of insects in the area was 15,000. The sc

Mila [183]

The estimated population, if the rate is a growth rate is 34696 populations

The estimated population, if the rate is a decline rate is 5657 populations

<h3> Exponential functions</h3>

The standard exponential equation is expressed as $y = ab^x$

where;

a is the intercept
b is the growth rate
x is the time taken

If the initial population is 15,000 with a rate of 15%, then;

a) If the rate is a growth rate, the estimated population will be expressed as;

$y = 15000(1+0.15)^{6}\\y=15000(1.15)^6\\y = 15000(2.313)\\y \approx 34,696$

The estimated population, if the rate is a decline rate is 34696 populations

b) If the rate is a decline rate, the estimated population will be expressed as;

$y = 15000(1-0.15)^{6}\\y=15000(0.85)^6\\y = 15000(0.37715)\\y \approx 5,657$

The estimated population, if the rate is a decline rate is 5657 populations

Learn more on exponential function here: brainly.com/question/12940982

6 0

2 years ago

The valve was tested on 270 engines and the mean pressure was 6.6 lbs/square inch. Assume the variance is known to be 0.49. If t

BlackZzzverrR [31]

Answer:

$z=\frac{6.6-6.5}{\frac{0.7}{\sqrt{270}}}=2.347$

The p value for this case would be given by"

$p_v =2*P(z>2.347)=0.0189$

For this case since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is significantly different from 6.5 lbs/square inch at 10% of significance. So then there is not enough evidence to conclude that the valve does not perform to the specifications

Step-by-step explanation:

Information given

$\bar X=6.6$ represent the sample mean

$s=\sqrt{0.49}= 0.7$ represent the population deviation

$n=270$ sample size

$\mu_o =6.5$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

z would represent the statistic

$p_v$ represent the p value for the test

Hypothesis to verify

We want to verify if the true mean for this case is equal to 6.5 lbs/square inch or not , the system of hypothesis would be:

Null hypothesis: $\mu= 6.5$

Alternative hypothesis: $\mu \neq 6.5$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing we got:

$z=\frac{6.6-6.5}{\frac{0.7}{\sqrt{270}}}=2.347$

The p value for this case would be given by"

$p_v =2*P(z>2.347)=0.0189$

For this case since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is significantly different from 6.5 lbs/square inch at 10% of significance. So then there is not enough evidence to conclude that the valve does not perform to the specifications

8 0

4 years ago

What is k(-4) equal to? K:m => 7+m^2 K(-4) = [1]