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Vlad1618 [11]
3 years ago
14

What are the botanical names of all flowers in the world​

Mathematics
2 answers:
taurus [48]3 years ago
7 0

Answer: Common Name   Scientific Name

  Lilac More about Lilac...     Syringa

Lily Lilium

Lotus Flower More about Lotus Flower... Nelumbo

Marigold More about Marigold... Tagetes

Step-by-step explanation:

1) Carnations. These ruffly flowers are perfect for the romantic on a budget. ...

2) Irises. Purple is the color of royalty, so it's no wonder that these flowers stand for faith and hope. ...

3) Lavender. This sweet-smelling flower is given to others as a sign of devotion. ...

4) Roses. ...

5) Tulips. ...

6) Sunflowers. ...

7) Gardenias. ...

8) Orchids.

IgorLugansk [536]3 years ago
7 0

Answer:

the names on the right are the botanical names

African lily= Agapanthus

Alpine thistle= Eryngium

Amaryllis= Hippeastrum

Amazon lily= Eucharis

Arum Lily= Zantedeschia

Baby’s breath= Gypsophila

Barberton= Daisy gerbera

Bell Flower/Canterbury Bells= Campanula

Bells of Ireland= Moluccella

Bird of paradise=Strelizia

Bleeding Heart= Dicentra spectabilis

Bloom= Chrysanthemum

Blue throatwort= Trachelium

Broom= Genista

Busy Lizzie= Impatiens

Calla lily= Zantedeschia

Canterbury Bells/Bell Flower= Campanula

Carnation= Dianthus

Chincerinchee Ornithogalum

Christmas rose Hellebore

Cockscomb Celosia

Columbine Aquilegia

Coneflower Echinacea

Cornflower Centaurea

Daffodil Narcissus

Evening primrose Oenothera

Feverfew Tanacetum parthenium

Flame tip Leucadendron

Flamingo flower/painter’s palette Anthurium

Forget-me-not Myosotis

Foxglove Digitalis

Gay feather Liatris

Globe thistle Echinops

Golden rod Solidago

Grape hyacinth Muscari

Guernsey lily Nerine

Hyacinth Hyacinthus

Iris Iris

Jersey lily Amaryllis Belladonna

Lady’s mantle  Alchemilla

Larkspur     Delphinium consolida

Lavender Lavandula

Lilac Syringa

Lily   Lilium

Lisianthus Eustoma

Lobster claw   Heliconia

Love in a mist   Nigella

Lupin Lupinus

Marigold  Calendula

Michaelmas Daisy   Aster

Mimosa    Acacia

Moth orchid Phelanopsis

Mums Spray chrysanthemum

Painter’s palette/Flamingo flower Anthurium

Peony Paeonia

Peruvian lily Alstromeria

Prairie gentian Lisianthus

Rose Rosa

Scabious  Scabiosa

Sea lavender   Limonium

September flower   Aster

Singapore orchid    Dendrobium

Snapdragon    Antirrhinum

Spray carnation  Dianthus

Statice Limonium

Stock Matthiola

Sugarbush Protea

Sunflower Helianthus

Sweet pea Lathyrus

Sweet william  Dianthus barbatus

Sword lily Gladiolus

Transvaal daisy Gerbera

Tulip Tulipa

Waxflower Chamaelaucium

Windflower Anemone

Yarrow Achillia

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At a restaurant, three salads, two sandwiches, and one drink cost $17.75. One salad, one sandwich, and three drinks cost $10.75.
Leno4ka [110]

Answer:

A salad costs $2.50

A sandwich costs $4.50

A drink costs $1.25

Step-by-step explanation:

Let x represent the salad, y represent the sandwich, and z represent the drink.

Since three salads, two sandwiches, and one drink cost $17.75:

3x + 2y + z = 17.75  (1)

Since one salad, one sandwich, and three drinks cost $10.75:

x + y + 3z = 10.75  (2)

Since a salad costs twice as much as a drink:

x = 2z  (3)

Multiply the equation 2 by -2 then, sum the equation 1 and equation 2:

-2x - 2y - 6z = -21.50

3x + 2y + z = 17.75

→ x - 5z = -3.75

Replace the x with 2z using equation 3:

2z - 5z = -3.75

-3z = -3.75

z = 1.25

x = 2z → x = 2.50

x + y + 3z = 10.75 → 2.50 + y + 3.75 = 10.75 → y + 6.25 = 10.75 → y = 4.50

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A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms
MissTica

Answer:

t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74

The degrees of freedom are given by:

df =n-1= 25-1=24

Now we can calculate the p value with the following probability:

p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}

And for this case since the p value is lower compared to the significance level \alpha=0.05 we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

Step-by-step explanation:

For this case we have the following info given:

\bar X = 39.5 represent the sample mean

s =6.6 represent the sample deviation

\mu = 30.6 represent the reference value to test.

n = 25 represent the sample size selected

The statistic for this case is given by:

t =\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

And replacing we got:

t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74

The degrees of freedom are given by:

df =n-1= 25-1=24

Now we can calculate the p value with the following probability:

p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}

And for this case since the p value is lower compared to the significance level \alpha=0.05 we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

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