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3 years ago

Please help !! what’s the answer? i don’t understand !!

Mathematics

1 answer:

dedylja [7]3 years ago

7 0

Answer:

992

Step-by-step explanation:

You must convert the meter the kilometer by miltiplying by 1000

and when you do it you must multiply 0.992 by 1000 since it is a fraction .

or you can work it this way :

0.992⇒1m
x(the new rate) ⇒1000m
x= 0.992*1000= 992

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Joshua wants to buy a flat-screen TV that costs $625. He saves $62.50 each week to buy the TV. In how many weeks will he have en

MissTica

Answer:

<h2><u><em>C. 10 weeks</em></u></h2>

Step-by-step explanation:

Joshua wants to buy a flat-screen TV that costs $625. He saves $62.50 each week to buy the TV. In how many weeks will he have enough money to make his purchase?

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3 years ago

1/4 K +1/4 M -2/3 K +5/9 M simplified

solniwko [45]

Answer:

-5/12k -11/36m

Step-by-step explanation:

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3 years ago

If 1 candy bar weighs 3 ounces what is the weight of 11 candy bars

Vera_Pavlovna [14]

Answer:

33 ounces

Step-by-step explanation:

Just multiply. If one weights 3 ounces 11x3=33

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4 years ago

Read 2 more answers

What is 3/8f+1/2=6(1/16f-3)

MA_775_DIABLO [31]

$\frac{3}{8f}+\frac{1}{2}=6*(\frac{1}{16f}-3) \\\\ \frac{3}{8f}^{(2}+\frac{1}{2}^{(8f}=\frac{6}{16f}-18^{(16f} \\\\ 6+8f=6-288f \\\\ 8f+288f=6-6 \\\\ 296f=0 \\\\ \boxed{f=0}$

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3 years ago

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Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago