What number is halfway between .2 and .3

Which of the following nonlinear inequalities is graphed below

Rashid [163]

Answer:

(x²/4) + (y²/16) ≤ 1

Step-by-step explanation:

The general form of the ellipse in the graph is

x²/4 + y²/16 = 1

we want to find the inequality which represents the shaded area this will determine whether which of the following 2 choices are correct

Choice 1: x²/4 + y²/16 ≥ 1

Choice 2: x²/4 + y²/16 ≤ 1

we note that point (0,0) lies inside the shaded area. Which means if we substitute (0,0), it should create an equality that is valid.

try choice 1: 0²/4 + 0²/16 = 0 ≥ 1 (not valid because obviously zero is smaller than 1)

try choice 1: 0²/4 + 0²/16 = 0 ≤ 1 (this is valid, hence choice 2 is the answer)

4 0

2 years ago

Oakville is a fungal disease that infects oak trees scientists have discovered that a single tree in a small forest is infected

Galina-37 [17]

Answer:

I solved part a

To solve this question, we need to solve an exponential equation, which we do applying the natural logarithm to both sides of the equation, getting that it will take 7.6 years for for 21 of the trees to become infected.

PART C

The logarithmic model is: g(x)= in x/0.4

We are given an exponential function, for the amount of infected trees f(x) after x years.To find the amount years needed for the number of infected trees to reach x, we find the inverse function, applying the natural logarithm.

Step-by-step explanation:

mark me brainliest!!

4 0

1 year ago

How to graph y = −4x − 1

Olin [163]

Answer: are you asking for this or

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

$14.40 for 4.5 pounds of beef

Svetradugi [14.3K]

Answer:

One pound of beef is $3.2

Step-by-step explanation:

14.40 / 4.5

= 3.2

CHECK:

3.2 * 4.5

=14.40

6 0

2 years ago

Read 2 more answers

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

2 years ago