This Question: 1 pt

What is the percent of change from 10 to 5?

NNADVOKAT [17]

Answer:

50% decrease

Step-by-step explanation:

$\dfrac{(V_2-V_1)}{|V_1|} \times 100$

$= \dfrac{(5 - 10)}{|10|} \times 100$

$= \dfrac{-5}{10} \times 100$

$= -0.5 \times 100$

$= -50\% \; \text{change}$

$= 50\% \; \text{decrease}$

8 0

3 years ago

Read 2 more answers

If (x+2) is a factor of x^3-6x^2+kx+10, k=

Alja [10]

If a binomial $(x-a)$ is a factor of a polynomial, then $a$ is a root of this polynomial.

$(x+2)=(x-(-2)) \Rightarrow a=-2\\\\
(-2)^3-6\cdot(-2)^2+k\cdot(-2)+10=0\\
-8-24-2k+10=0\\
-2k=22\\\
\boxed{k=-11}
$

3 0

3 years ago

PLSSS HELP IMEDEATLY Find the x- and y-intercepts of the graph of 2 – 6y = 35. State your answers as

Lelechka [254]

Answer:

Y= -5.5 There is no X intercept.

Step-by-step explanation:

Please mark brainliest.

7 0

3 years ago

A tent has the dimensions shown in the diagram. Find the area of the groundsheet needed to fit inside the tent. Round your answe

kogti [31]

Apologies if the diagram is a bit confusing

7 0

3 years ago

] It is claimed that 42% of US college graduates had a mentor in college. For a sample of college graduates in Colorado, it was

Bad White [126]

Answer:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

Step-by-step explanation:

Information given

n=1045 represent the random sample selected

X=502 represent the college graduates with a mentor

$\hat p=\frac{502}{1045}=0.480$ estimated proportion of college graduates with a mentor

$p_o=0.42$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is higher than 0.42, the system of hypothesis are.:

Null hypothesis: $p \leq 0.42$

Alternative hypothesis: $p > 0.42$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info we got:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

7 0

3 years ago