All categories

3 years ago

. What is the area of a rectangle with a length of 5/8 foot and a width of 4/9 foot? *

Mathematics

1 answer:

ankoles [38]3 years ago

5 0

Answer:

B. 20/72 sq ft

Step-by-step explanation:

5/8(4/9) = 20/72 ft² = 5/18 ft²

You might be interested in

What does the shaded part represent in the given Venn -diagram

Snezhnost [94]

Pls look at your question before asking a question. You didn’t give us a venn diagram to look at.

Brainliest?

5 0

3 years ago

Read 2 more answers

A is a special relation where one input is related with exactly one output

Solnce55 [7]

Answer:

function

A function is a relation (such that for each input, there is exactly one output) between sets. The formula and the graph are merely descriptions of this relation.

6 0

3 years ago

Find the measure of the missing angles

serg [7]

Answer:

y = 57° and x = 33°

Step-by-step explanation:

I've attached a picture.

:)

5 0

2 years ago

if 2/3 of the girls in class have brown eyes and 1/4 of the girls have blue eyes, what fraction of the girls in class have neith

Evgesh-ka [11]

1 - ( 2/3 + 1 /4 ) =
= 1 - ( 8 / 12 + 3 / 12 ) =
= 1 - 11 / 12 = 12 / 12 - 11 /12 = 1 /12
Answer:
1 / 12 of the girls in class have neither blue or brown eyes.

4 0

4 years ago

The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of sixte

goldenfox [79]

Answer:

Yes, it can be concluded that the mean hexane concentration is less in treated water than in unsaturated water

Step-by-step explanation:

The number of of specimen in the samples of untreated water, n₁ = 16

The sample mean, $\overline x_1$ = 228.0

The sample standard deviation, s₁ = 4.3

The number of of specimen in the samples of treated water, n₂ = 20

The sample mean, $\overline x_2$ = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; $\overline x_1$ ≥ $\overline x_2$

The alternative hypothesis, Hₐ; $\overline x_1$ < $\overline x_2$

The degrees of freedom = 16 - 1 = 15

The test statistic, $t_{\alpha}$ = 1.341

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Plugging in the values, we get;

$t=\dfrac{(224.6- 228.0)}{\sqrt{\dfrac{5.0^{2}}{20} -\dfrac{4.3^{2} }{16}}} \approx -11.0675$

Given that the t-value is large, the corresponding p-value is low, therefore, we fail to reject the null hypothesis and there is considerable statistical evidence to suggest that the mean hexane concentration is less in treated than in untreated water, therefore, we have; $\overline x_1$ ≥ $\overline x_2$

6 0

3 years ago