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2 years ago

PLEASE HELP, I REALLY NEED IT!!!

Mathematics

1 answer:

kogti [31]2 years ago

3 0

The number which is express in each of the models as given in the image attached to the task content are as follows;

a). 1.37

b). 1.37

c). 1.37.

<h3>What numbers are expressed according to the given models in the task content?</h3>

It follows from the task content that the models describe that One flat represents 1 whole, One rod represents 1 tenth and one unit represents 1 hundredth.

It therefore follows from the task content that in each of the models, the algebraic sum of flat(s), rods and units as the case may be results in the value; 1.37 as the utmost number represented by the models.

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Please Help! Geometry!

tresset_1 [31]

Step-by-step explanation:

At first we can find AC

then we can find m<A and m< C

or, m<C and then m< A

m<B is already given in the question so there is no need of finding it.

7 0

3 years ago

The number of bacteria present in a culture after t minutes is given as where k is a constant. if there are 8520 bacteria presen

andrew-mc [135]

Given that the situation has been modeled by the formula:
B=1000e^(kt)
given that B=8520 after t=15 min
then the value of k will be solved as follows:
plugging in the values in the formula and solving for k
8520=1000e^(15k)
8.52=e^(15k)
solving for k
15k=ln(8.52)
hence
k=ln(8.52)/15
k=0.1428
k=0.143

8 0

3 years ago

Read 2 more answers

If someone could help me I would really appreciate it!<br><br> Factor by slide and divide!

OleMash [197]

Answer:

(3x-4)(2x+5)

Step-by-step explanation:

6x^2 + 7x + 24

= 6x^2 + (15-8)x + 24

= 6x^2 + 15x - 8x + 24

= 3x( 2x+5 ) - 4( 2x+5 )

= (3x-4)(2x+5)

5 0

3 years ago

A study reveals that 65% of homeowners in Omaha own a lawn mower. To test this claim, you form a sample of 500 homeowners in Oma

kotegsom [21]

Answer:

There is no enough evidence to reject the claim using a level of significance of 5%

Step-by-step explanation:

Let p be the true proportion of homeowners in Omaha who own a lawnmower. We want to test

$H_{0}: p = p_{0} = 0.65$ vs $H_{1}: p\neq 0.65$ (two-tailed alternative).

We have a large sample of size n = 500 and a point estimate for p is $\hat{p}=0.68$ . The test statistic is given by

$Z=\frac{\hat{p}-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}=\frac{0.68-0.65}{\sqrt{0.65(1-0.65)/500}}$ where $\sqrt{0.65(1-0.65)/500}$ is the standard error of $\hat{p}$ when $H_{0}$ is true. The observed value is $z_{0}$ = 1.4064 which comes from a standard normal distribution. Because we want to use a level of significance of 5%, our rejection region is RR={z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles respectively. Because 1.4064 does not belong to RR, we fail to reject the null hypothesis.

6 0

3 years ago

Will mark Brainliest

Ivahew [28]

Answer:

57.39

[tex]440 \: ( 1 + \frac{5t}{46 + \: {t}^{2} } ) \\ 440 \: ( \frac{6t}{46 {t}^{2} } ) \: ( \: multiply \: 440 \times 6t) \\ \frac{2640}{46} \\ = 57.39 or 57.30

8 0

3 years ago