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3 years ago

1 hundreth is how many times greater than 1 thousandths? Pls helppp

Mathematics

2 answers:

Verdich [7]3 years ago

4 0

Answer:

1 hundreths is 10 times greater than 1 thousandths.

Step-by-step explanation:

just like 100 is ten times greater than 10.

frez [133]3 years ago

3 0

Answer:

1 hundredth is 10x greater than 1 thousandth.

Step-by-step explanation:

1/1000 * 10 = 10/1000

10/1000 = 1/100

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PLEASE HELP ME ASAP I WILL MARK YOU AS BRILLIANT please help with these three

GalinKa [24]

Answer:

26: x=34

28: x=17

30: m=30

Step-by-step explanation:

26: 2x+22=90

subtract 22

2x=68

divide by 2

x=34

28: 18+3x+21=90

add 18 and 21

3x+39=90

subtract 39

3x=51

divide by 3

x=17

30: 43+87+m+20=180

add 43 and 87 and 20

m+150=180

subtract 150

m=30

6 0

3 years ago

You have discovered another alien. Find the correct rescue path before answering this question. Which statements are true about

IrinaVladis [17]

Answer:

(a) The y-intercept is not visible on the graph.

(D) The line crosses the x-axis at –3

Step-by-step explanation:

i already did this on edge

8 0

3 years ago

Read 2 more answers

Hailey signed up for a streaming music service that costs $14 per month. The service allows Hailey to listen to unlimited music,

liraira [26]

Answer:

The total is: 74$

Step-by-step explanation:

1.50$ times 40 equals 60. Then you have to add up the cost you usually get per month which is 14$. 60 plus 14$ = 74$

8 0

3 years ago

90. The number of daily evening newspapers in the United States has decreased by 46.5% to

aleksandr82 [10.1K]

Answer:

700

Step-by-step explanation:

Given:

Daily evening newspapers in 2015 = 389

Percentage decrease from 2000 to 2015 = 46.5%

To find:

Number of newspapers in 2000 = ?

Solution:

Let the number of newspapers in 2000 = $x$

When 46.5% of the total newspapers in 2000 ( $x$ ) is subtracted from the total newspapers in 2000 ( $x$ ), it will be equal to the number of newspapers in 2015.

Making the equation as per the given statement:

$x-x\times 46.5\%=389\\\Rightarrow \dfrac{100x-46.5x}{100}=389\\\Rightarrow 53.5x=38900\\\Rightarrow x\approx \bold{727}$

Therefore, the answer is:

There were 727 newspapers in 2000.

7 0

3 years ago

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran

mart [117]

Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

$X=48$ number of vehicles classified as sports utility.

$n=500$ random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

$p$ true population proportion of vehicles classified as sports utility vehicles.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

The confidence interval would be given by this formula :

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

7 0

3 years ago