The number of significant figures in 0.01500 is

Answer the question is it <br> A <br> B <br> C or <br> D

MrRa [10]

i think its c lmk if i am wrong

7 0

3 years ago

If you change the sign of a point’s

zubka84 [21]

It will reflect the same distance over the Y axis, from the right side of the graph to the left side of the graph.

4 0

3 years ago

Gavin had $5460.40 that he divided into tow bank accounts. he deposited 3/8 into a savings account. he deposited the rest into h

Ipatiy [6.2K]

(5460.4)(3/8) = 2,047.65 to savings

5460.4 - 2,047.65 = 3,412.75 checking

4 0

3 years ago

An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t

erastova [34]

Answer:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

$\bar X=46.5$ represent the mean

$\sigma=1.1$ represent the population standard deviation

$n=150$ sample size

$\mu_o =46.7$ represent the value to verify

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:

Null hypothesis: $\mu = 46.7$

Alternative hypothesis: $\mu \neq 46.7$

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing we got:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

7 0

3 years ago

Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents

Dmitry [639]

Answer:

Option A. $f(x)=86[1.04]^{x}$ ; grows approximately at a rate of 0.4% daily

Step-by-step explanation:

we have

$f(x)=86(1.29)^{x}$

where

f(x) the number of weeds in the garden

x ----> the number of weeks

Calculate how quickly the weeds grow each day

Remember that a week is equal to seven days

so

$f(x)=86(1.29)^{\frac{x}{7}}$

Using the law of exponents

b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

so

$f(x)=86[(1.29)^{\frac{1}{7}}]^{x}$

$f(x)=86[1.04]^{x}$

therefore

The rate is approximately

1.04=1+r

r=1.04-1=0.04=4% daily

8 0

3 years ago