All categories

3 years ago

Reuben is making a replica of the US flag for his school project.

1 answer:

8 0

A) The perimeter is twice the sum of length and width.
.. 2*(2.5 cm +4.75 cm) = 14.5 cm

b) (1 in)/(0.5 cm) = (flag perimeter)/(14.5 cm)
.. (14.5 cm)*(1 in)/(.5 cm) = flag perimeter
.. flag perimeter = 29 in

You might be interested in

Liam has 5 kg of plant food. If he uses 25 g of the food on each plant, how many plants can he feed?

bearhunter [10]

Answer:

He can feed 200 plants

Step-by-step explanation:

To solve this, we will follow the step below:

first, convert 25 g to kg

25 g = 25/1000 = 0.025 kg

Now using proportion to solve;

Let x be the number of food plant he can feed the plant

0.025 kg = 1

5 kg = x

cross-multiply

0.025×x = 5×1

0.025x =5

Divide both-side of the equation by 0.025

0.025x/0.025 = 5/0.025

x=200 plants

Therefore, he can feed 200 plants

3 0

2 years ago

How do you solve this one 10x^2-25=x^2

solong [7]

10x^2 - 25 = x^2
Take the square root of each side.

$\sqrt{x^2} = \sqrt{10x^2 - 25}$
x = 10x - 5i

4 0

2 years ago

(1.5 x 10^2) X (2.0 x 10^3) Express your answer in scientific notation and standard form.

zavuch27 [327]

Answer:

3*10^5 or 300,000 cant remember witch one it is

Step-by-step explanation:

1.5*10^2= 150

2*10^3=2,000

150*2000=300,000

3*10^5

8 0

2 years ago

Pls awnser this question I will give brainliest pls

Makovka662 [10]

Answer:

My guess is G

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In o

Mandarinka [93]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 22 ounces

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

Let $\mu$ = average ounces of passion fruit juice used to fill each bottle.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 22 ounces {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight of the passion fruit juice = 21.54 ounces

$\sigma$ = population standard deviation = 1.38 ounce

n = sample of bottles = 65

So, test statistics = $\frac{21.54-22}{\frac{1.38}{\sqrt{65} } }$

= -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.

6 0

3 years ago